poj 2689

A的挺费劲的。。。用筛法求某个区间内的素数。

思路想通的挺快就是实现的真难。。

//============================================================================
// Name        : 2689.cpp
// Author      :
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

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

bool tag[60000], flag[1000010];
long long l, r, Max, Min, t0, pre, t, num;
long long prime[1000010];


int main() {
	freopen("a.txt", "r", stdin);
    num = 0;
    tag[1] = true;
    for(int i = 2;i <= 60000;i++){
        if(tag[i] == false){
            prime[++num] = i;
            for(int j = i<<1;j <= 60000;j+=i){
                tag[j] = true;
            }
        }
    }
	while(scanf("%d%d", &l, &r)!=EOF){
	    if(l == 1) l = 2;
		memset(flag, false, sizeof(flag));
		for(int i = 1;prime[i]*prime[i] <= r;i++){
            t = l/prime[i]+(l%prime[i]!=0);
            if(t == 1) t++;
            for(long long j = t*prime[i];j <= r;j+=prime[i]){
                flag[j-l] = true;
            }
		}
		t0 = 0;
		Max = 0;
		Min = 2147483647;
		for(long long i = l;i <= r;i++){
			if(flag[i-l] == false&&t0 == 0){
				t0 = 1;
				pre = i;
			}
			else if(flag[i-l] == false&&t0 == 1){
				t0 = 2;
				t = i-pre;
				if(Max < t) Max = t;
				if(Min > t) Min = t;
				pre = i;
			}
			else if(flag[i-l] == false){
				t = i-pre;
				if(Max < t) Max = t;
				if(Min > t) Min = t;
				pre = i;
			}
		}
		if(t0 != 2){
			printf("There are no adjacent primes.\n");
		}
		else{
//		    printf("%lld, %lld\n", Min, Max);
			int t1 = 0;
			for(long long i = l;i <= r;i++){
				if(flag[i-l] == false&&t1 == 0){
					t1 = 1;
					pre = i;
				}
				else if(flag[i-l] == false&&t1 == 1){
					t1 = 2;
					t = i-pre;
					if(t == Min){
						printf("%lld,%lld are closest, ", pre, i);
						break;
					}
					pre = i;
				}
				else if(flag[i-l] == false){
					t = i-pre;
					if(t == Min){
						printf("%lld,%lld are closest, ", pre, i);
						break;
					}pre = i;
				}
			}
			t1 = 0;
			for(long long i = l;i <= r;i++){
				if(flag[i-l] == false&&t1 == 0){
					t1 = 1;
					pre = i;
				}
				else if(flag[i-l] == false&&t1 == 1){
					t1 = 2;
					t = i-pre;
					if(t == Max){
						printf("%lld,%lld are most distant.\n", pre, i);
						break;
					}
					pre = i;
				}
				else if(flag[i-l] == false){
					t = i-pre;
					if(t == Max){
						printf("%lld,%lld are most distant.\n", pre, i);
						break;
					}
					pre = i;
				}
			}
		}
	}
	return 0;
}