2404 Super Prime（欧拉筛素数）

2404 Super Prime（欧拉筛素数）

Problem Description

We all know, prime is a kind of special number which has no other factors except of 1 and itself.
2,3,5,7,11,13,17,19,23,29 are the top 20 primes.
Now there is a new question about prime:
We call a prime as super prime when and only when which can be represented as the sum of multiple continuous primes. For example: 5=2+3, so 5 is a super prime.
Please program to judge whether a number is a super prime or not.

Input

The first line is an integer T (T<=1000), and then T lines follow.
There is only one positive integer N(1

Output

For each case you should output the case number first, and then the word “yes” if the integer N is a super prime, and you should output “no” otherwise.

Sample Input

3
5
7
8

Sample Output

Case 1: yes
Case 2: no
Case 3: no

#include<iostream>
#include<algorithm>
#include<memory.h>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
const int M=100000;
int prim[M];
bool vis[M];
int prime(int n) {//欧拉筛素数O（n）
    int cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2; i<n; i++) {
        if(!vis[i]) {
            prim[cnt++]=i;
        }
        for(int j=0; j<cnt&&i*prim[j]<n; j++) {
            vis[i*prim[j]]=1;
            if(i%prim[j]==0) {
                break;
            }
        }
    }
    return cnt;
}
int look(int num,int tol) {//二分查找
    int mid=0;
    int low=0;
    int high=num;
    while(low<high) {
        mid=(low+high)/2;
        if(tol==prim[mid]) {
            return mid;
        } else if(tol<prim[mid]) {
            high=mid;
        } else if(tol>prim[mid]) {
            low=mid+1;
        }
    }
    return -1;
}
int main() {
    int n;
    cin>>n;
    int x;
    int num=prime(M);
    for(int i=1; i<=n; i++) {
        cin>>x;
        cout<<"Case "<<i<<": ";
        int flag=0;
        int inx=look(num,x);
        /*for(int j=0;j<num;j++){//线性查找也可以A
            if(prim[j]==x){
                flag=1;
                inx=j;
                break;
            }
        }*/
        if(inx!=-1) {
            int flags=0;
            for(int j=0; j<inx; j++) {//连续的几个质数相加，这个很重要
                int sum=0;
                for(int k=j;k<inx;k++){
                    sum+=prim[k];
                    if(sum>x){
                        break;
                    }
                    else if(sum==x){
                        flags=1;
                        break;
                    }
                }
                if(flags==1){
                    break;
                }
            }
        if(flags==0) {
            cout<<"no"<<endl;
        } else {
            cout<<"yes"<<endl;
        }
    }
    else {
        cout<<"no"<<endl;
    }
}
return 0;
}


/***************************************************
User name: dlili
Result: Accepted
Take time: 76ms
Take Memory: 344KB
Submit time: 2018-04-13 10:29:22
****************************************************/