2014/4/28 多校第九次

C：快速求N以内因数和，N以内互质数的和。

容斥版：

#include <iostream>
#include <stdio.h>
#include <string.h>
#define maxn 1100000
#define LL long long
//N以内gcd(i,N)==1的i的和
using namespace std;
bool flag[maxn];
int prim[maxn/3],cnt;
int pri[209],exp[209],count;

void built(){
    cnt=0;
    memset(flag,0,sizeof(flag));
    for(int i=2;i<maxn;i++){
         if(!flag[i]){
               prim[cnt++]=i;
               for(int j=2;j*i<maxn;j++) flag[j*i]=true;
         }
    }
//    for(int i=0;i<100;i++) cout<<prim[i]<<" ";cout<<endl;
}
void toprim(LL n){
    int i=0;
    count=0;
    memset(exp,0,sizeof(exp));
    while(i<cnt && prim[i]<=n){
        if(n%prim[i]==0){
            pri[count]=prim[i];
            while(n%prim[i]==0) {
                n=n/prim[i];
                exp[count]++;
            }
            count++;
        }
        i++;
    }
    if (n>1) {
        exp[count]=1;
        pri[count++]=n;
    }
//    cout<<"count="<<count<<endl;
}
LL sigm(LL x){
    return  (1+x)*x/2;
}
LL Z(LL x,LL n){
    return x*sigm(n/x);
}
LL solveB(LL n){
    LL ans=0;
    for(int i=1;i<(LL)(1<<count);i++){
        int times=0;
        LL mul=1;
        for(int j=0;j<count;j++){
            if(i&((LL)(1<<j))) {
                times++;mul*=pri[j];
            }
        }
        if (times%2) {
            ans+=Z(mul,n);
        }else {
            ans-=Z(mul,n);
        }
    }
    return ans;
}
LL ans=0;
LL getexp(LL p,int k){
    LL ans=1;
    while(k--) ans*=p;
    return ans;
}
void dfs(LL n,LL t,int k){
    if (k==count) {
        if (t<=n) ans+=t;
        return ;
    }
    for (int i=0;i<=exp[k];i++){
        dfs(n,t*getexp((LL)pri[k],i),k+1);
    }
    return ;
}
LL solveA(LL n){//返回k*i==n,i<=n,sigm(i)
    ans=0;
    dfs(n,1,0);
    return ans;
}
int N;
int main()
{
//    freopen("out.txt","w",stdout);
    built();
    while(cin>>N){
//    for(LL N=1;N<=300;N++){
        toprim(N);
        LL A=solveA(N);
        LL B=sigm(N)-solveB(N);
        printf("%lld
",A-B);
    }
    return 0;
}

公式版：

H:

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <stack>
#include <vector>
#include <map>
#include <queue>
#include <stdlib.h>
#define LL long long
#define INF 999999999
#define eps 0.00000001
#define maxn  1010
using namespace std;
int mat[maxn][maxn],up[maxn][maxn],lef[maxn][maxn],rig[maxn][maxn];
int main(){
    int N,M;
    while(~scanf("%d%d",&M,&N)){
            swap(N,M);
            for(int i=0;i<M;i++)
                for(int j=0;j<N;j++)
                    scanf("%d",&mat[i][j]);
            LL ans=0;
            for(int i=0;i<M;i++){
                int lo=-1,ro=N;
                for(int j=0;j<N;j++){
                    if (mat[i][j]==0) {
                            up[i][j]=lef[i][j]=0;
                            lo=j;
                    }
                    else {
                        if (i==0) up[i][j]=1;else up[i][j]=up[i-1][j]+1;
                        if (i==0) lef[i][j]=lo+1;else lef[i][j]=max(lef[i-1][j],lo+1);
                    }
                }
                for(int j=N-1;j>=0;j--){
                    if (mat[i][j]==0){
                        rig[i][j]=N;
                        ro=j;
                    }else{

                        if (i==0) rig[i][j]=ro-1;else rig[i][j]=min(rig[i-1][j], ro-1);
                        int m=min(up[i][j],(rig[i][j]-lef[i][j]+1));
                        if (m<=0) continue;
                        ans=max(ans,(LL)m*(LL)m);
                    }
                }
            }

            printf("%lld
",ans);
    }
    return 0;
}