zoukankan      html  css  js  c++  java
  • 2018 Multi-University Training Contest 1

    1001 Maximum Multiple

    谈学姐写的

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 long long n,i,j,k,s,t,ans,T;
     4 long long a[1000010];
     5 int main()
     6 {
     7         n=1e6;
     8         for (i=1;i<=n;i++)
     9         {
    10                 t=-1;
    11                 if (i%3==0)
    12                 {
    13                         s=i/3;
    14                         t=s*s;
    15                         t=t*s;
    16                 }
    17                 if (i%4==0)
    18                 {
    19                         s=i/4;
    20                         ans=s*s;
    21                         ans=ans*2*s;
    22                         if (ans>t) t=ans;
    23                 }
    24                 if (i%6==0)
    25                 {
    26                         s=i/6;
    27                         ans=s*s*2;
    28                         ans=ans*s*3;
    29                         if (ans>t) t=ans;
    30                 }
    31                 a[i]=t;
    32         }
    33         scanf("%lld",&T);
    34         while (T--)
    35         {
    36                 scanf("%lld",&n);
    37                 printf("%lld
    ",a[n]);
    38         }
    39 }
    谈学姐

    1002 Balanced Sequence

    瞎比排序

      1 #include<bits/stdc++.h>
      2 using namespace std;
      3 struct sam
      4 {
      5         long long x,y,z;
      6 };
      7 long long n,i,j,k,s,t,ans,T;
      8 sam a[100010];
      9 int b[100010];
     10 char ch;
     11 int cmp(sam &p, sam &q)
     12 {
     13         return (p.x<q.x||(p.x==q.x&&p.y>q.y));
     14 }
     15 int cmp2(sam &p, sam &q)
     16 {
     17         return (p.y>q.y||(p.x>q.x&&p.y==q.y));
     18 }
     19 int main()
     20 {
     21         scanf("%lld",&T);
     22         while (T--)
     23         {
     24                 scanf("%lld",&n);
     25                 ch=getchar();
     26                 ans=0;
     27                 for (i=1;i<=n;i++)
     28                 {
     29                         ch=getchar();
     30                         t=0;
     31                         while (ch!='
    ')
     32                         {
     33                                 t++;
     34                                 if (ch=='(') b[t]=0;
     35                                 else
     36                                 {
     37                                         b[t]=1;
     38                                         if (t>=2&&b[t-1]==0)
     39                                         {
     40                                                 t-=2;
     41                                                 ans++;
     42                                         }
     43                                 }
     44                                 ch=getchar();
     45                         }
     46                         s=1;
     47                         while (s<=t&&b[s]==1) s++;
     48                         s--;
     49                         a[i].x=s;a[i].y=t-s;a[i].z=0;
     50                 }
     51                 sort(a+1,a+1+n,cmp);
     52                 s=0;
     53                 for (i=1;i<=n;i++)
     54                 {
     55                         if (a[i].x<a[i].y)
     56                         {
     57                                 if (s>=a[i].x)
     58                                 {
     59                                         ans+=a[i].x;
     60                                         s=s-a[i].x+a[i].y;
     61                                 }
     62                                 else
     63                                 {
     64                                         ans+=s;
     65                                         s=a[i].y;
     66                                 }
     67                                 a[i].z=1;
     68                         }
     69                 }
     70                 for (i=1;i<=n;i++)
     71                 {
     72                         if (a[i].x==a[i].y)
     73                         {
     74                                 if (s>=a[i].x)
     75                                 {
     76                                         ans+=a[i].x;
     77                                 }
     78                                 else
     79                                 {
     80                                         ans+=s;
     81                                         s=a[i].y;
     82                                 }
     83                                 a[i].z=1;
     84                         }
     85                 }
     86                 sort(a+1,a+1+n,cmp2);
     87                 for (i=1;i<=n;i++)
     88                 {
     89                         if (a[i].x>a[i].y)
     90                         {
     91                                 if (s>=a[i].x)
     92                                 {
     93                                         ans+=a[i].x;
     94                                         s=s-a[i].x+a[i].y;
     95                                 }
     96                                 else
     97                                 {
     98                                         ans+=s;
     99                                         s=a[i].y;
    100                                 }
    101                                 a[i].z=1;
    102                         }
    103                 }
    104                 ans*=2;
    105                 printf("%lld
    ",ans);
    106         }
    107 }
    谈学姐

    1003 Triangle Partition

    谈学姐写的

    #include<bits/stdc++.h>
    using namespace std;
    struct sam
    {
            long long x,y,z;
    };
    long long n,i,j,k,s,t,ans,T;
    sam a[4010];
    int cmp(sam &p, sam &q)
    {
            return (p.x<q.x);
    }
    int main()
    {
            scanf("%lld",&T);
            while (T--)
            {
                    scanf("%lld",&n);
                    for (i=1;i<=3*n;i++)
                    {
                            scanf("%lld%lld",&a[i].x,&a[i].y);
                            a[i].z=i;
                    }
                    sort(a+1,a+1+3*n,cmp);
                    t=0;
                    for (i=1;i<=n;i++)
                    {
                            printf("%lld %lld %lld
    ",a[t+1].z,a[t+2].z,a[t+3].z);
                            t+=3;
                    }
            }
    }
    谈学姐

    1004 Distinct Values

    区间排序后从前往后扫的时候把前面不属于本区间的暴力删掉用set维护没用过的值

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 1e5 + 10;
     4 int a[maxn], b[maxn];
     5 typedef pair<int, int> pii;
     6 vector<pii> v;
     7 set<int> NOT, USED;
     8 set<int> :: iterator it;
     9 
    10 int main() {
    11     int T;
    12     scanf("%d", &T);
    13     while(T--) {
    14         int n, m;
    15         scanf("%d %d", &n, &m);
    16         NOT.clear(), USED.clear();
    17         for(int i = 1; i <= n; ++i) b[i] = 0, NOT.insert(i);
    18         v.clear();
    19         for(int i = 1; i <= m; ++i) {
    20             int l, r;
    21             scanf("%d %d", &l, &r);
    22             v.push_back(pii(l, r));
    23         }
    24         sort(v.begin(), v.end());
    25         int lst = 1, p = 1;
    26         for(int i = 0; i < m; ++i) {
    27             int l = v[i].first, r = v[i].second;
    28             while (lst < l) {
    29                 if(USED.find(b[lst]) != USED.end()) USED.erase(b[lst]), NOT.insert(b[lst]);
    30                 lst++;
    31             }
    32             p = max(p, l);
    33             while(p <= r) {
    34                 it = NOT.begin();
    35                 b[p++] = *it;
    36                 NOT.erase(b[p-1]);
    37                 USED.insert(b[p-1]);
    38             }
    39         }
    40         for(int i = 1; i <= n; ++i) if(!b[i]) b[i] = 1;
    41         for(int i = 1; i <= n; ++i) printf("%d%c", b[i], i == n ? '
    ' : ' ');
    42     }
    43     return 0;
    44 }
    Aguin

    1005 Maximum Weighted Matching

    按边打牌,用f[u][v]表示两端取/不取的最大值,g[u][v]表示对应的方案数

    首先把所有重边合并,然后每次选择一个度为2的点将两条串联的边合并

    由于一开始把重边都去了,这时合并两条串联边的时候可能会出现至多一条边和它并联,例如考虑一个三元环的情况,这时把两条并联边也合并

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const LL mod = 1e9 + 7;
     5 const int maxn = 1e5 + 10;
     6 
     7 struct e {
     8     int u, v;
     9     LL f[2][2], g[2][2];
    10     e(int U = -1, int V = -1, int W = -1) {
    11         u = U, v = V, memset(f, -1, sizeof(f)); memset(g, 0, sizeof(g));
    12         if(W != -1) f[0][0] = 0, f[1][1] = W, g[0][0] = g[1][1] = 1;
    13     }
    14     bool operator < (const e & o) const {return v < o.v;}
    15 };
    16 set<e> G[maxn];
    17 
    18 inline void ud(LL & f, LL & g, LL nf, LL ng) {
    19     if(nf > f) f = nf, g = ng;
    20     else if(nf == f) g = (g + ng) % mod;
    21 }
    22 
    23 queue<int> q;
    24 int main() {
    25     int T;
    26     scanf("%d", &T);
    27     while(T--) {
    28         int n, m, y = -1;
    29         scanf("%d %d", &n, &m);
    30         for(int i = 1; i <= n; ++i) G[i].clear();
    31         for(int i = 1; i <= m; ++i) {
    32             int u, v, w;
    33             scanf("%d %d %d", &u, &v, &w);
    34             set<e> :: iterator it = G[u].find(e(0, v));
    35             if(it == G[u].end()) G[u].insert(e(u, v, w)), G[v].insert(e(v, u, w));
    36             else if((*it).f[1][1] == w) {
    37                 e t = *it; t.g[1][1]++; G[u].erase(t); G[u].insert(t);
    38                 t = *(G[v].find(e(0, u))); t.g[1][1]++; G[v].erase(t); G[v].insert(t);
    39             }
    40             else if((*it).f[1][1] < w){
    41                 e t = *it; t.f[1][1] = w; t.g[1][1] = 1; G[u].erase(t); G[u].insert(t);
    42                 t = *(G[v].find(e(0, u))); t.f[1][1] = w; t.g[1][1] = 1; G[v].erase(t); G[v].insert(t);
    43             }
    44         }
    45         while(!q.empty()) q.pop();
    46         for(int i = 1; i <= n; ++i) if(G[i].size() <= 2) q.push(i);
    47         while(!q.empty()) {
    48             int x = q.front(); q.pop();
    49             if(G[x].size() == 1) {y = x; continue;}
    50             e a = *G[x].begin(), b = *G[x].rbegin(), c(a.v, b.v); G[x].clear();
    51             G[a.v].erase(e(0, x)), G[b.v].erase(e(0, x));
    52             for(int ua = 0; ua <= 1; ++ua) for(int va = 0; va <= 1; ++va)
    53                 for(int ub = 0; ua + ub <= 1; ++ub) for(int vb = 0; vb <= 1; ++vb)
    54                     if(a.f[ua][va] == -1 || b.f[ub][vb] == -1) continue;
    55                     else ud(c.f[va][vb], c.g[va][vb], a.f[ua][va] + b.f[ub][vb], a.g[ua][va] * b.g[ub][vb] % mod);
    56             if(G[a.v].find(e(0, b.v)) != G[a.v].end()) {
    57                 e d = *G[a.v].find(e(0, b.v)), f = e(a.v, b.v);
    58                 G[a.v].erase(e(0, b.v)), G[b.v].erase(e(0, a.v));
    59                 if(G[a.v].size() == 1) q.push(a.v);
    60                 if(G[b.v].size() == 1) q.push(b.v);
    61                 for(int ua = 0; ua <= 1; ++ua) for(int va = 0; va <= 1; ++va)
    62                     for(int ub = 0; ua + ub <= 1; ++ub) for(int vb = 0; va + vb <= 1; ++vb)
    63                         if(c.f[ua][va] == -1 || d.f[ub][vb] == -1) continue;
    64                         else ud(f.f[ua + ub][va + vb], f.g[ua + ub][va + vb], c.f[ua][va] + d.f[ub][vb], c.g[ua][va] * d.g[ub][vb] % mod);
    65                 c = f;
    66             }
    67             G[a.v].insert(c);
    68             swap(c.u, c.v), swap(c.f[1][0], c.f[0][1]), swap(c.g[1][0], c.g[0][1]);
    69             G[b.v].insert(c);
    70         }
    71         e a = *G[y].begin();
    72         LL ans1 = -1, ans2 = 0;
    73         for(int ua = 0; ua <= 1; ++ua) for(int va = 0; va <= 1; ++va)
    74             if(a.f[ua][va] != -1) ud(ans1, ans2, a.f[ua][va], a.g[ua][va]);
    75         printf("%lld %lld
    ", ans1, ans2);
    76     }
    77     return 0;
    78 }
    Aguin

    1006 Period Sequence

    杜老师教的做法~

    问题等价于求$(x, i, j)$三元组满足$s[i] = s[j] = x$

    若$i = j$,贡献是$(i - a + 1)(b - i + 1)$,若$i < j$,贡献是$(i - a + 1)(b - j + 1)$,$i > j$和前面一样直接乘二就好

    考虑$i = j$的情况,首先枚举每个余数$r$,把$0$至$n - 1$中每个余$r$的项抠出来,记这些位置为$pos$,每个位置对应着一个等差数列

    对于每个位置对应的等差数列,找到它为$r$的那一项在原数列$S$中的位置,并记这个位置为这个等差数列的第$0$项

    那么这个等差数列的每一项在原数列中的位置都是它在自己所在的等差数列中的位置的一次函数,而等差数列本身就是一次函数

    于是所要求的贡献其实是一个三次函数的前缀和,也就是四次函数,求个前五项,用拉格朗日插出来,然后找到$[a, b]$对应等差数列中项数的位置,减一减

    对于$i < j$的情况,枚举两个等差数列,只有当他们相同的项同时在$[a, b]$中间才有贡献,同样是四次函数

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const LL mod = 1e9 + 7;
     5 LL L[2222], R[2222];
     6 int s[2222];
     7 
     8 // length, f[1, ..., n], f[x]
     9 LL inv_fac[2222], pre[2222], suf[2222];
    10 LL Lagrange(int n, LL *fx, LL x) {
    11     if(1 <= x && x <= n) return fx[x];
    12     inv_fac[0] = inv_fac[1] = 1;
    13     for(int i = 2; i <= n; ++i) inv_fac[i] = inv_fac[mod % i] * (mod - mod / i) % mod;
    14     for(int i = 3; i <= n; ++i) inv_fac[i] = inv_fac[i - 1] * inv_fac[i] % mod;
    15     pre[0] = suf[n + 1] = 1;
    16     for(int i = 1; i <= n; ++i) pre[i] = pre[i - 1] * (x % mod + mod + mod - i) % mod;
    17     for(int i = n; i >= 1; --i) suf[i] = suf[i + 1] * (x % mod + mod + mod - i) % mod;
    18     LL ret = 0;
    19     for(int i = 1; i <= n; ++i) {
    20         LL t = pre[i - 1] * suf[i + 1] % mod * inv_fac[i - 1] % mod * inv_fac[n - i] % mod;
    21         if((n - i) % 2) t = t * (mod - 1) % mod;
    22         ret = (ret + t * fx[i]) % mod;
    23     }
    24     return ret;
    25 }
    26 
    27 int main() {
    28     int T;
    29     scanf("%d", &T);
    30     while(T--) {
    31         int n;
    32         LL a, b, ans = 0;
    33         scanf("%d %lld %lld", &n, &a, &b);
    34         for(int i = 0; i < n; ++i) scanf("%d", s + i);
    35         for(int r = 0; r < n; ++r) {
    36             vector<int> pos;
    37             for(int i = 0; i < n; ++i)
    38                 if(s[i] % n == r) pos.push_back(i - s[i] + r);
    39             sort(pos.begin(), pos.end());
    40             for(int i = 0; i < pos.size(); ++i) {
    41                 LL sum = 0, fx[6];
    42                 L[i] = (a - pos[i] + n - 1) / n;
    43                 while(pos[i] + n * L[i] < a) L[i]++;
    44                 while(pos[i] + n * (L[i] - 1) >= a) L[i]--;
    45                 R[i] = (b - pos[i]) / n;
    46                 while(pos[i] + n * (R[i] + 1) <= b) R[i]++;
    47                 while(pos[i] + n * R[i] > b) R[i]--;
    48                 for(int k = 0; k < 5; ++k) {
    49                     sum = (sum + (r + k * n) * (pos[i] + k * n + mod - a % mod + mod + 1) % mod * (b % mod - pos[i] - k * n + mod + 1)) % mod;
    50                     fx[k + 1] = sum;
    51                 }
    52                 ans = (ans + Lagrange(5, fx, R[i] + 1) - Lagrange(5, fx, L[i]) + mod) % mod;
    53             }
    54             for(int i = 0; i < pos.size(); ++i) {
    55                 for(int j = i + 1; j < pos.size(); ++j) {
    56                     LL ml = max(L[i], L[j]), mr = min(R[i], R[j]);
    57                     if(ml > mr) continue;
    58                     LL sum = 0, fx[6];
    59                     for(int k = 0; k < 5; ++k) {
    60                         sum = (sum + (r + k * n) * (pos[i] + k * n + mod - a % mod + mod + 1) % mod * (b % mod - pos[j] - k * n + mod + 1)) % mod;
    61                         fx[k + 1] = sum;
    62                     }
    63                     ans = (ans + 2 * (Lagrange(5, fx, mr + 1) - Lagrange(5, fx, ml) + mod)) % mod;
    64                 }
    65             }
    66         }
    67         printf("%lld
    ", ans);
    68     }
    69     return 0;
    70 }
    Aguin

    1007 Chiaki Sequence Revisited

    谈学姐写的

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 struct sam
     4 {
     5         long long x,y,z;
     6 };
     7 long long n,i,j,k,s,t,ans,T,l,r,x,mid,temp,temp2,ni,mod=1e9+7,temp3;
     8 long long a[100010];
     9 int main()
    10 {
    11         scanf("%lld",&T);
    12         ni=(1+mod)/2;
    13 
    14         while (T--)
    15         {
    16                 scanf("%lld",&n);
    17                 if (n<=2)
    18                 {
    19                         printf("%lld
    ",n);
    20                         continue;
    21                 }
    22                 ans=1;n--;
    23                 l=2;r=n;x=0;
    24                 while (l<=r)
    25                 {
    26                         mid=(l+r)/2;
    27                         s=0;t=mid;
    28                         while (t>0)
    29                         {
    30                                 s+=t;
    31                                 t=t/2;
    32                         }
    33                         if (s>=n)
    34                         {
    35                                 x=mid;
    36                                 r=mid-1;
    37                         } else l=mid+1;
    38                 }
    39                 t=x-1;
    40                 s=0;
    41                 while (t>0)
    42                 {
    43                         s+=t;
    44                         t=t/2;
    45                 }
    46                 t=x-1;temp3=1;
    47                 while (t>0)
    48                 {
    49                         temp=(1+t)%mod;
    50                         temp2=t%mod;
    51                         temp=temp*temp2%mod;
    52                         temp=temp*ni%mod;
    53                         temp=temp*temp3%mod;
    54                         ans=(ans+temp)%mod;
    55                         t=t/2;
    56                         temp3=temp3*2%mod;
    57                 }
    58                 temp=(n-s)%mod;
    59                 x=x%mod;
    60                 temp=temp*x%mod;
    61                 ans=(ans+temp)%mod;
    62                 while (ans<0) ans+=mod;
    63                 printf("%lld
    ",ans);
    64         }
    65 }
    谈学姐

    1008 RMQ Similar Sequence

    比赛的时候没有发现跨过最大值两边就没影响了……

    建笛卡尔树的时候实际上单调栈维护的是最右的一条链

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const LL mod = 1e9 + 7;
     5 const int maxn = 1e6 + 10;
     6 int a[maxn], st[maxn], ls[maxn], rs[maxn], vis[maxn];
     7 LL ans, inv[maxn];
     8 
     9 int sz[maxn];
    10 void dfs(int x) {
    11     sz[x] = 1;
    12     if(ls[x]) dfs(ls[x]), sz[x] += sz[ls[x]];
    13     if(rs[x]) dfs(rs[x]), sz[x] += sz[rs[x]];
    14     ans = ans * inv[sz[x]] % mod;
    15 }
    16 
    17 int main() {
    18     inv[1] = 1;
    19     for(int i = 2; i < maxn; ++i) inv[i] = inv[mod % i] * (mod - mod / i) % mod;
    20     int T;
    21     scanf("%d", &T);
    22     while(T--) {
    23         int n, p = 0;
    24         scanf("%d", &n);
    25         for(int i = 1; i <= n; ++i) ls[i] = rs[i] = vis[i] = 0;
    26         for(int i = 1; i <= n; ++i) {
    27             scanf("%d", a + i);
    28             int pop = 0;
    29             while(p && a[st[p]] < a[i]) pop = 1, p--;
    30             if(p) rs[st[p]] = i;
    31             if(pop) ls[i] = st[p+1];
    32             st[++p] = i;
    33         }
    34         for(int i = 1; i <= n; ++i) vis[ls[i]] = vis[rs[i]] = 1;
    35         for(int i = 1; i <= n; ++i) if(!vis[i]) ans = 1, dfs(i);
    36         printf("%lld
    ", ans * n % mod * inv[2] % mod);
    37     }
    38     return 0;
    39 }
    Aguin

    1009 Lyndon Substring

    论文不学

    1010 Turn Off The Light

    $f[i][0/1]$表示$1$至$i - 1$全为$0$,$i$的状态为$0/1$,$i + 1$至$n$为原串, 从$i$开始往右关完的最少步
    $g[i]$表示从$i$出发把左边关完回到i的最少步,$h[i]$表示从i出发把左边关完回到$i$后位置$i$的$0/1$状态
    终点可能是最右边的灯位置$l$或者往左一个位置$l - 1$

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const LL mod = 1e9 + 7;
     5 const int maxn = 1e6 + 10;
     6 const int INF = 1e9;
     7 int f[maxn][2], g[maxn], h[maxn];
     8 int n, b[maxn], z[maxn];
     9 char s[maxn];
    10 
    11 void solve() {
    12     int l = n + 1, r = 0;
    13     for(int i = 1; i <= n; ++i) if(b[i]) l = min(l, i), r = max(r, i);
    14     if(l > r) {
    15         for(int i = 1; i <= n; ++i) z[i] = 0;
    16         return;
    17     }
    18     f[r][0] = 0, f[r][1] = 3;
    19     if(1 < r) f[r - 1][0] = 1, f[r - 1][1] = 2;
    20     for(int i = r - 2; i >= 1; --i) {
    21         f[i][0] = f[i + 1][b[i + 1] ^ 1] + 1;
    22         f[i][1] = f[i + 1][b[i + 1]] + 3;
    23     }
    24     g[l + 1] = 2, h[l + 1] = b[l + 1] ^ 1;
    25     for(int i = l + 2; i <= r; ++i) {
    26         if(h[i - 1]) g[i] = g[i - 1] + 2, h[i] = b[i] ^ 1;
    27         else g[i] = g[i - 1] + 4, h[i] = b[i];
    28     }
    29     for(int i = 1; i <= l; ++i) z[i] = min(z[i], f[i][b[i]]);
    30     for(int i = l + 1; i <= r; ++i) z[i] = min(z[i], g[i] + f[i][h[i]]);
    31 }
    32 
    33 int main() {
    34     int T;
    35     scanf("%d", &T);
    36     while(T--) {
    37         scanf("%d %s", &n, s + 1);
    38         for(int i = 1; i <= n; ++i) b[i] = s[i] - '0', z[i] = INF;
    39         solve();
    40         reverse(b + 1, b + 1 + n), reverse(z + 1, z + 1 + n);
    41         solve();
    42         reverse(z + 1, z + 1 + n);
    43         LL ans = 0;
    44         for(int i = 1; i <= n; ++i) ans = (ans + (LL) i * z[i]) % mod;
    45         printf("%lld
    ", ans);
    46     }
    47     return 0;
    48 }
    Aguin

    1011 Time Zone

    沙雕题还WA一发

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 
     4 int main() {
     5     int N;
     6     scanf("%d", &N);
     7     while(N--) {
     8         int h, m;
     9         char s[11];
    10         scanf("%d %d %s", &h, &m, s);
    11         int l = strlen(s), dh, dm;
    12         if(l == 5) {
    13             if(s[3] == '+') dh = 8 - (s[4] - '0'), dm = 0;
    14             else dh = 8 + (s[4] - '0'), dm = 0;
    15         }
    16         else if(l == 6) {
    17             if(s[3] == '+') dh = 8 - ((s[4] - '0') * 10 + (s[5] - '0')), dm = 0;
    18             else dh = 8 + ((s[4] - '0') * 10 + (s[5] - '0')), dm = 0;
    19         }
    20         else if(l == 7) {
    21             if(s[3] == '+') dh = 8 - (s[4] - '0'), dm = -6 * (s[6] - '0');
    22             else dh = 8 + (s[4] - '0'), dm = 6 * (s[6] - '0');
    23         }
    24         else {
    25             if(s[3] == '+') dh = 8 - ((s[4] - '0') * 10 + (s[5] - '0')), dm = -6 * (s[7] - '0');
    26             else dh = 8 + ((s[4] - '0') * 10 + (s[5] - '0')), dm = 6 * (s[7] - '0');
    27         }
    28         h -= dh, m -= dm;
    29         while(m >= 60) m -= 60, h++;
    30         while(m < 0) m += 60, h--;
    31         while(h >= 24) h -= 24;
    32         while(h < 0) h += 24;
    33         printf("%02d:%02d
    ", h, m);
    34     }
    35     return 0;
    36 }
    Aguin
  • 相关阅读:
    Redis数据类型有哪些?
    python---判断元素是否可用 is_displayed/is_enabled/is_selected
    pythone---获取文本链接,text/get_attribute/current_url/title
    python---清空 clear
    python---获取元素 CSS
    JDK安装及配hi环境
    python---获取元素 Xpath
    python---获取元素 id/name/class_name/link_text/partial选择器及注意事项
    python---数据类型----set集合和字典
    web搭建自动化环境
  • 原文地址:https://www.cnblogs.com/Aguin/p/9364408.html
Copyright © 2011-2022 走看看