SGU 107
题意：输入一个N，表示N位数字里面有多少个的平方数的结尾9位是987654321
收获：打表，你发现相同位数的数相乘结果的最后几位，就和那两个相乘的数最后几位相乘一样，比如3416*8516 = 29090656，它的最后两位就和16*16=256的最后两位一样 为56，那么你发现987654321位9位，而且你预处理出的那8个答案就是9位，你就看d=n-9为多少位，那么就是9*8*（d位10）相乘，一个for就行了
#include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<b;++i) typedef long long ll; int main(){ //for(ll i =2;i<=1e9;++i) if(i*i%mod==x) cout<<i<<" "; int n; scanf("%d",&n); if(n<9) return puts("0"),0; if(n==9) return printf("8"),0; int d = n - 9; printf("72"); rep(i,1,d) printf("0"); return 0; }
SGU 104
题意：给你n多花，i<j的话，i花只能放在j花前面的花瓶，每个花和对应花瓶有不同的魅力值，问你怎么放花使得魅力值最大
收获：一开始想错了，看到了数据很小，而且很小匹配的问题，没有注意到花只能按顺序放，然后就写了最小费用流（边值弄成魅力值的相反数），然后一直re在2
最后看了题解，看到dp，就很快想出了转移方程，然后初始化这些的又搞了很久，真菜~~，dp挺好理解的，直接看代码吧
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e2+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int c[maxn][maxn],dp[maxn][maxn]={0}; int path[maxn][maxn]; vector<int> ans; int main(){ int n,m; scanf("%d%d",&n,&m); rep(i,0,n+1)rep(j,0,m+1) dp[i][j]=-inf; rep(i,1,n+1) rep(j,1,m+1) scanf("%d",&c[i][j]); dp[0][0]=0; rep(i,1,n+1)rep(j,i,m+1)rep(k,i-1,j){ if(dp[i][j]<dp[i-1][k]+c[i][j]){ dp[i][j]=dp[i-1][k]+c[i][j]; path[i][j] = k; } } int x,y; int mx = -inf; rep(i,n,m+1) if(mx<dp[n][i]){ mx = dp[n][i]; x=n;y=i; } printf("%d ",mx); while(x!=1){ ans.pb(y); y=path[x][y];x--; } ans.pb(y); reverse(all(ans)); rep(i,0,sz(ans)) printf("%d%c",ans[i]," "[i+1==sz(ans)]); return 0; }
SGU 127
题意：做个电话簿，前两页目录，然后一页最多k个电话，电话首位不同的不能在同一页
收获：无
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 8025; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int a[maxn],cnt[12]={0}; int main(){ int k,n,ans=2; scanf("%d%d",&k,&n); rep(i,0,n) scanf("%d",a+i); sort(a,a+n); rep(i,0,n) cnt[a[i]/1000]++; rep(i,0,10) if(cnt[i]){ ans += (cnt[i]+k-1)/k; } printf("%d ",ans); return 0; }