View Code of Problem 120

#include <stdio.h>
int main(){
    int i,j,su[60000],n1,n2,m1,m2,ans,k;
    char c;
    for(i=2;i<60000;i++){
        su[i]=1;
    }
    su[0]=0,su[1]=0;
    for(i=1;i<60000;i++){
        if(su[i]==1){
            for(j=i*2;j<60000;j+=i){//求解质数
                su[j]=0;
            }
        }
    }
    while(scanf("%d%c%d",&n1,&c,&m1)!=EOF){
        ans=0;
        scanf("%d%c%d",&n2,&c,&m2);
        while(n1<n2){
            while(m1<60){
                k=n1*2500+m1;
                if(su[k]==1){
                    ans++;
                }
                    m1++;
            }
                m1=0;
                n1++;
        }
            while(m1<=m2){
                k=n1*2500+m1;
                if(su[k]==1){
                    ans++;
                }
                m1++;
            }
            printf("%d\n",ans);
    }
    return 0;
}