『算法-ACM 竞赛-POJ』 3581 Prime Gap（二分）

Prime Gap
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 11009 Accepted: 6298
Description
The sequence of n − 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, ‹24, 25, 26, 27, 28› between 23 and 29 is a prime gap of length 6.
Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.
Input
The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.
Output
The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.
Sample Input
10 11 27 2 4921700
Sample Output
4 0 6 0 114
Source Japan 2007

二分，稍微需要处理一点细节。

#include<iostream>
#include<cstdio>
using namespace std;
const int MAXN=1e7-100;
int prime[MAXN];
bool pri[MAXN+10];
int sieve(int n)
{
    int p=0;
    for(int i=0; i<=n; i++)
        pri[i]=true;
    pri[0]=pri[1]=false;
    for(int i=2; i<=n; i++)
    {
        if(pri[i])
        {
            prime[++p]=i;
            for(int j=2*i;j<=n;j+=i)
                pri[j]=false;
        }
    }
    return p;
}
int main()
{
    int m=sieve(1e7-100);
    int t;
    while(~scanf("%d",&t)&&t){
        int up=m;
     //   cout<<prime[m];
        int lo=0;
        while(1){
            int tem=(up+lo)/2;
            if(prime[tem]>t) up=tem;
            else if(prime[tem]==t){
                up=lo=0;
                break;
            }
            else lo=tem;
            if(lo+1==up)break;
        }
       printf("%d\n",prime[up]-prime[lo]);
    }
}