2020牛客寒假算法基础训练营2

2020牛客寒假算法基础训练营2

A.做游戏(思维)

传送门

题意：两个人玩石头剪刀布，牛牛出石头a次，剪刀b次，布c次，牛可乐出石头x次，剪刀y次，布z次，问牛牛最多可以赢多少次

题解：对牛牛每次可以赢的出法取最小值即可

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>
 
using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
 
int main()
{
    ll a,b,c,x,y,z;
    cin>>a>>b>>c>>x>>y>>z;
    ll ans = min(a,y)+min(b,z)+min(c,x);
    cout<<ans<<endl;
    return "BT7274", NULL;
}
 
inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

B.排数字(思维)

传送门

题意：给定一个由纯数字构成的字符串，你可以随机排列这些数字，问你排出的字符串最多包含多少个“616”子串，61616这样的算两个

题解：最好的情况肯定是“61616”这样排列，结果就是6和1的数量取最小值。

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();

int main()
{
    int n,cnt6 = 0,cnt1 = 0;
    string s;
    cin>>n>>s;
    for(int i = 0; i < n; i++){
        if(s[i]=='6')cnt6++;
        if(s[i]=='1')cnt1++;
    }
    cout<<min(cnt6-1,cnt1)<<endl;
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

C.算概率(DP)

传送门

题意：一共n道题，每道题做对的概率为pi，问做对0-n道题的概率分别是多少，结果对1e9+7取模

题解：(dp_{i,j})表示前i道题做对了j道，则有状态转移方程(dp_{i,j}) = (dp_{i-1,j})x(1-(p_i))+(dp_{i-1,j-1})x(p_i)，注意j=0的特殊情况，时间复杂度O((n^2))

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 2019;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
ll n,p[N],dp[N][N];
const ll mod = 1e9+7;
int main()
{
    cin>>n;
    dp[0][0] = 1;
    for(int i = 1; i <= n; i++)
        cin>>p[i];
    for(int i = 1; i <= n; i++){
        dp[i][0] = dp[i-1][0]*(mod+1-p[i])%mod;
        for(int j = 1; j <= i; j++){
            dp[i][j] = (dp[i-1][j]*(mod+1-p[i])+dp[i-1][j-1]*p[i])%mod;
        }
    }
    for(int i = 0; i <= n; i++){
        cout<<dp[n][i]<<" ";
    }
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

D.数三角(极角排序)

传送门

题意：给定n个不重复的点，求一共可以组成多少个钝角三角形

题解：对每个点进行极角排序后双指针找上下界统计钝角数量

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
struct node1
{
    double x,y;
}a[1000];
double node[1500];
int main()
{
    int n;
    cin>>n;
    for(int i = 1; i <= n; i++)cin>>a[i].x>>a[i].y;
    ll ans = 0;
    for(int i = 1; i <= n; i++){
        int cnt = 0;
        for(int j = 1; j <= n; j++){
            if(i==j)continue;
            node[++cnt] = atan2(a[i].y-a[j].y,a[i].x-a[j].x);
            if(node[cnt] < 0)node[cnt]+=2*PI;
        }
        sort(node+1,node+1+cnt);
        for(int j = 1; j <= cnt; j++) node[j+cnt] = node[j]+2*PI;
        int l = 1,r = 1;
        for(int j = 1; j <= cnt; j++){
            while(r <= 2*cnt&&node[r] - node[j] < PI)r++;
            while(l <= 2*cnt&&node[l] - node[j] <= 0.5*PI)l++;
            ans+=r-l;
        }
    }
    cout<<ans<<endl;
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

E.做计数(完全平方数)

传送门

题意：给定一个n，问有多少个不同的三元组{i，j，k}，满足(sqrt{i}+sqrt{j} = sqrt{k})，i，j，k均为正整数，且i * j <= n，i，j，k有一者不同即为不同

题解：展开算式可得 i + j + 2(sqrt{ij}) = k，可见i*j必须是完全平方数，枚举n以内的完全平方数，再枚举其因数即可

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();

int main()
{
    int ans = 0;
    int n;
    cin>>n;
    for(int i = 1; i*i <= n; i++){
        for(int j = 1; j <= i; j++){
            if(i*i%j==0){
                ans+=2;
            }
        }ans-=1;
    }
    cout<<ans<<endl;
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

F.拿物品(贪心)

传送门

题意：对于n个物品，每个物品都有一个a值和b值，牛牛和牛可乐依次选择这n个物品，每次从剩下的里面选一个，牛牛先选，牛牛最后的权值为选择物品的a值总和，牛可乐的权值为选择的物品的b值总和，两者都希望自己的权值最大，在两者的最优选择方案下两者会如何选择

题解：贪心，两者都尽量选择a+b值大的物品，可以理解成削弱你就是增强我、

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
struct node
{
    int a,b,v,id;
};

bool cmp(node a,node b)
{
    return a.v>b.v;
}
int main()
{
    int n;
    cin>>n;
    node a[N];
    for(int i = 1; i <= n; i++){
        cin>>a[i].a;
    }
    for(int i = 1; i <= n; i++){
        cin>>a[i].b;
        a[i].v = a[i].a+a[i].b;
        a[i].id = i;
    }
    vector<int> v1,v2;
    sort(a+1,a+n+1,cmp);
    for(int i = 1; i <= n; i++){
        if(i&1){
            v1.push_back(a[i].id);
        }else
            v2.push_back(a[i].id);
    }
    for(int i = 0; i < v1.size(); i++){
        cout<<v1[i]<<" ";
    }cout<<endl;
    for(int i = 0; i < v2.size(); i++){
        cout<<v2[i]<<" ";
    }cout<<endl;
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

G.判正误(hash)

传送门

题意：给定t组a，b，c，d，e，f，g，a-g的范围在int内，判断(a^d + b^e + c^f = g)是否成立

题解：直接计算肯定是会TLE的，这里我选择了快速幂%1e9+7这个神奇的数字居然过了，跟出题人所描述的方法不太相同，也不太理解这道题目的意义

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>
 
using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
const ll mod = 1e9+7;
ll quick_mod(ll a,ll b)
{
    ll ans = 1,t = a;
    while(b){
        if(b&1==1){
            ans = (ans*t)%mod;
        }
        b>>=1;
        t = (t*t)%mod;
    }
    return ans;
}
int main()
{
    Test{
        ll a,b,c,d,e,f,g,x,y,z,ans;
        cin>>a>>b>>c>>d>>e>>f>>g;
        x = quick_mod(a,d);
        y = quick_mod(b,e);
        z = quick_mod(c,f);
        ans = x+y+z;
        if(g==ans){
            cout<<"Yes"<<endl;
        }else
            cout<<"No"<<endl;
         
    };
    return "BT7274", NULL;
}
 
inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

H.施魔法(DP)

传送门

题意：一共n个元素，第i个元素的值为ai，每次从中最少选择k个元素，费用是每次选中的元素的极差，每个元素只能选择一次，求最小花费

题解：先从小到大排序，选择方案一定是连续的一系列数，并且选择次数越多花费越小，假设将这个数列切割m次，每次切割都会使最大花费减少切割位置左右两个元素的差值，比如在i位置切割，花费会减少(a_{i+1} - a_i)，这样即可得到状态转移方程，dp[i] = (max_{jin[1,i-k]}){dp[j]}+d[i]，dp[i]表示在位置i处切割后最多减少的花费，d[i]表示在i位置切割减少的花费，其中最大值可以在计算中维护，时间复杂度O(n)

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
int a[N],d[N],dp[N];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i = 1; i <= n; i++){
        cin>>a[i];
    }
    sort(a+1,a+1+n);
    ll ans = a[n] - a[1];
    int now = 0;
    for(int i = 1; i <= n-1; i++){
        d[i] = a[i+1] - a[i];
    }
    memset(dp,0,sizeof(dp));
    int pre = 0;
    for(int i = k; i <= n-k; i++){
        pre = max(pre,dp[i-k]);
        dp[i] = pre+d[i];
        now = max(now, dp[i]);
    }
    cout<<ans-now<<endl;
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

I.建通道(二进制)

传送门

题意：给定n个点，每个点都有一个权值v，两点之间建立一条路的花费为lowbit(v1(igoplus)v2)，问连通所有点的最小花费

题解：将权值相同的点去重，设去重后有m个点，接着利用&和|运算统计出所有权值中二进制中既有1又有0的位数，找到最小的二进制第k位，结果就是(2^k*(m-1))

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>
 
using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
int v[N];
int main()
{
    int v1,v0,n;
    cin>>n;
    v1 = 0x7fffffff,v0 = 0;
    for(int i = 1; i <= n; i++){
        cin>>v[i];
        v1 &= v[i];
        v0 |= v[i];
    }
    v1 ^= v0;
    sort(v+1,v+1+n);
    int m = 0;
    for(int i = n; i > 0; i--){
        if(v[i]!=v[i+1])m++;
    }
    ll ans;
    for(int i = 0; i <= 30; i++){
        int cur = 1<<i;
        if(cur & v1){
            ans = 1ll*cur*(m-1);
            break;
        }
    }
    cout<<ans<<endl;
    return "BT7274", NULL;
}
 
inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}

J.求函数(线段树)

传送门

题意：有n个一次函数，第i个函数f(i) = (k_i)*x+(b_i)，有m次操作，每次操作分两种

1 i k b 将(f_i(x))修改为(f_i(x) = k*x+b)

2 l r 求(f_r(f_{r-1}(...f_{l+1}(f_l(1))...)))，答案对1e9+7取模

题解：对于第二种操作所求结果可以计算为

[prod_{i=l}^rk_i+sum_{i=l}^rb_i*prod_{j=i+1}^rk_j ]

需要注意的使当j=i+1>r时按(prod_{j=i+1}^rk_j) = 1计算。

那么对于上式，我们可以采用两棵线段树分别维护加号左右的两个式子，重点就是如何合并这两个区间

对于区间[l,r]，我们把它分成两部分([l_1,r_1])，([l_2,r_2])，令式子(prod_{i=l_1}^{r_1}k_i = n1)，(prod_{i=l_2}^{r_2}k_i = n2)

(sum_{i=l_1}^rb_i*prod_{j=i+1}^rk_j) = m1，(sum_{i=l_2}^rb_i*prod_{j=i+1}^rk_j) = m2，那么合并区间时，加号左边部分的合并方法就是 (n1*n2)，加号右侧的合并方法是(m1*n2+m2)，注意这些式子累加符号的上限仍然是r

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;
using ll = long long;
const ll N = 2e5+5;
const ll mod = 1e9+7;
using pii = pair<int,int>;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
ll st1[N<<2],st2[N<<2],k[N];
void modify(int id, int l, int r, int i, int k, int b)
{
    if(l == r){
        st1[id] = k;st2[id] = b;
        return ;
    }
    int mid = (l+r)>>1;
    if(i<=mid) modify(id<<1,l,mid,i,k,b);
    else
        modify(id<<1|1,mid+1,r,i,k,b);
    st1[id] = (st1[id<<1]*1ll*st1[id<<1|1])%mod;
    st2[id] = (st2[id<<1]*1ll*st1[id<<1|1]+st2[id<<1|1])%mod;
}
pii merge( pii a, pii b)
{
    return {(a.first*1ll*b.first)%mod,(a.second*1ll*b.first+b.second)%mod};
}
pii query(int id,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return {st1[id],st2[id]};
    int mid = (l+r)>>1;
    if(qr<=mid){
        return query(id<<1,l,mid,ql,qr);
    }
    if(ql>mid){
        return  query(id<<1|1,mid+1,r,ql,qr);
    }
    return merge(query(id<<1,l,mid,ql,qr),query(id<<1|1,mid+1,r,ql,qr));
}
int main()
{
    int n,m,b;
    cin>>n>>m;
    for(int i = 1; i <= n; i++)cin>>k[i];
    for(int i = 1; i <= n; i++){
        cin>>b;
        modify(1,1,n,i,k[i],b);
    }
    while(m--){
        int f;
        cin>>f;
        if(f==1){
            int i,k,b;
            cin>>i>>k>>b;
            modify(1,1,n,i,k,b);
        }else{
            int l,r;
            cin>>l>>r;
            pii ans = query(1,1,n,l,r);
            cout<<(ans.first+ans.second)%mod<<endl;
        }
    }
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}