Prime Gap |
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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。
代码语言：javascript复制#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("%dn",prime[up]-prime[lo]);
}
}
