POJ 2739 Sum of Consecutive Prime Numbers【素数打表】
解题思路:给定一个数,判定它由几个连续的素数构成,输出这样的种数
用的筛法素数打表
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20020 | Accepted: 10966 |
Description
Input
Output
Sample Input
2
3
17
41
20
666
12
53
0
Sample Output
1
1
2
3
0
0
1
2
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int prime[10000],num[10005];
int main()
{
int i,j,k=0,tmp,n;
for(i=0;i<=10000;i++) num[i]=1;
num[0]=0;
num[1]=0;
for(i=2;i<=10000;i++)
{
if(num[i])
{
prime[k++]=i;
for(tmp=i*2;tmp<=10000;tmp+=i)
num[tmp]=0;
}
}
while(scanf("%d",&n)!=EOF&&n)
{
int ans=0,sum;
for(i=0;i<k;i++)
{
sum=0;
for(j=i;j<k&&sum<n;j++)
sum+=prime[j];
if(sum==n)
ans++;
}
printf("%d\n",ans);
}
}
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