Bi-shoe and Phi-shoe

解题思路：

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long ll;
int unprime[1000200];

void is_prime(){
    memset(unprime,sizeof(unprime));
    unprime[1] = 1;
    for(int i = 2; i <= 1000100; i++){//注意临界值
        int tt = 1000100/i;
        for(int j = 2; j <= tt; j++){
            unprime[i*j] = 1;
        }
    }
}

int main(){
    is_prime();
    int T,t = 1;
    scanf("%d",&T);
    while(T--){
        int n,x;
        scanf("%d",&n);
        ll ans = 0;
        for(int i = 0; i < n; i++){
            scanf("%d",&x);
            for(int j = x+1; ; j++){
                if(!unprime[j]){
                    ans += j;
                    break;
                }
            }
        }
        printf("Case %d: %lld Xukhan",t++,ans);
    }
    return 0;
}