bzoj 4173 打表???

没有任何思路,打表发现ans=phi(n)*phi(m)*n*m
%%% popoqqq Orz
然而并没有看懂……

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define mod 998244353ll
using namespace std;
long long getphi(long long x){
    long long ans=x;
    for(long long i=2;i*i<=x;i++)
        if(x%i==0){
            ans/=i;ans*=i-1;
            while(x%i==0)x/=i;
        }
    if(x!=1){ans/=x;ans*=x-1;}
    return ans;
}
int main(){
    long long  n,m,x,y,ans=1;
    scanf("%lld%lld",&n,&m);
    x=getphi(n)%mod;
    y=getphi(m)%mod;
    n%=mod; m%=mod;
    ans=n;ans=(ans*m)%mod;
    ans=(ans*x)%mod;ans=(ans*y)%mod;
    printf("%lld
",ans);
    return 0;
}

附打表程序

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 500
using namespace std;
int phi[N],prime[N],tot;
bool bo[N];
void init(){
    phi[1]=1;
    for(int i=2;i<=N;i++){
        if(!bo[i]){
            prime[++tot]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=tot&&i*prime[j]<=N;j++){
            bo[i*prime[j]]=1;
            if(i%prime[j]==0){
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
}
void work(int x,int y){
    int cnt=0;
    printf("phi[%d]==%d  phi[%d]==%d
",x,phi[x],y,phi[y]);
    for(int i=2;i<=x+y;i++)
        if(x%i+y%i>=i){
            printf("phi[%d]==%d
",i,phi[i]);
            cnt+=phi[i];
        }
    printf("%d  %d  %d  %d  ans==%d
",x,y,phi[x],phi[y],phi[x]*phi[y]*cnt);
}
int main(){
    int x,y;
    init();
    while(scanf("%d%d",&x,&y)==2)
        work(x,y);
}
原文地址:https://www.cnblogs.com/Ren-Ivan/p/7746695.html