【题解】质数距离

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题目描述

题目描述
给定两个整数 ，求闭区间$[L,R]$中相邻两个质数差值最小的数对与差值最大的数对。当存在多个时，输出靠前的素数对。

输入格式
多组数据。每行两个数$L,R$。
输出格式
详见输出样例。

样例

样例输入
2 17
14 17
样例输出
2,3 are closest, 7,11 are most distant.
There are no adjacent primes.

题目分析

这一道题很明显是素数筛的题，一般做素数筛都用线筛，因为要快一些，这道题也一样。

直接每一次用线筛，筛出在$[L,R]$这个区间的所有质数，然后再进行枚举比较求出间隔最小与间隔最大的相邻质数。

错误代码

#include 
#include 
#include 
#include 
using namespace std;
const int M = 1e8 + 5;
bool flag[M];
int pre[M];
int prime(int l, int r) {
    memset(flag, false, sizeof(flag));
    memset(pre, 0, sizeof(pre));
    int k = 0;
    if (l == 1) {
        l++;
    }
    for (int i = 2; i <= r; i++) {
        if (!flag[i]) {
            k++;
            pre[k] = i;
        }
        for (int j = 1; j <= k && pre[j] * i <= r; j++) {
            flag[pre[j] * i] = true;
            if (i % pre[j] == 0)
                break;
        }
    }
    return k;
    //	for(int i=1;i<=k;i++){
    //		printf("%d ",pre[i]);
    //	}
}
int main() {
    int L, R;
    while (scanf("%d %d", &L, &R) != EOF) {
        int len = prime(L, R);
        if (len == 1) {
            printf("There are no adjacent primes.\n");
        }
        int x1 = 0, x2 = 0, y1 = 0, y2 = 0;
        int minn = 0x3f3f3f3f, maxn = 0;
        for (int i = 1; i <= len - 1; i++) {
            if (pre[i + 1] - pre[i] < minn && pre[i] >= L) {
                x1 = i, x2 = i + 1;
                minn = pre[i + 1] - pre[i];
            }
            if (pre[i + 1] - pre[i] > maxn && pre[i] >= L) {
                y1 = i, y2 = i + 1;
                maxn = pre[i + 1] - pre[i];
            }
        }
        if (x1 == 0 && x2 == 0 && y1 == 0 && y2 == 0) {
            printf("There are no adjacent primes.\n");
            continue;
        }
        printf("%d,%d are closest, %d,%d are most distant.\n", pre[x1], pre[x2], pre[y1], pre[y2]);
    }
    return 0;
}

但是一提交就直接超时爆零了。

之后我就反思为何会超时。首先，主函数里遍历答案这是不可去除的；之后就是线筛函数也不会错。最后，我发现在每次输入$L,R$之后，都有一次线筛，那万一第一个数据都包含了之后的数据，就不是浪费时间了。

所以我们可以找到数据的极值，即$M$。在每次输入数据之前都进行一次最大的线筛，则后面的数据就直接使用就可以了。

正确代码

#include
#include
#include
using namespace std;
const int M=1e7+5;
bool flag[M],flag1[M];
int pre[M],arr[M];
int k;
void Prime2(){
	flag[0]=flag[1]=true;
	for(int i=2;i<=M;i++){
		if(flag[i]==false){
			pre[k++]=i;
		}
		for(int j=1;j<k&&pre[j]*i<M;j++){
			flag[pre[j]*i]=true;
		}
	}
}
int main(){
	int L,R;
	Prime2();
	while(scanf("%d %d",&L,&R)!=EOF){
		if(L==1){
			L=2;
		}
		memset(flag1,false,sizeof(flag1));
		for(int i=0;i<k;i++){
			int a=(L-1)/pre[i]+1;
			int b=R/pre[i];
			for(int j=a;j<=b;j++){
				if(j>1){
					flag1[pre[i]*j-L]=true;
				}
			}
		}
		int x1=0,x2=0,y1=0,y2=0;
		int minn=0x3f3f3f3f,maxn=-1,p=-1;
		for(int i=0;i<=R-L;i++){
			if(flag1[i]==false){
				if(p==-1){
					p=i;
					continue;
				}
				if(i-p<minn){
					x1=p+L,x2=i+L;
					minn=i-p;
				}
				if(i-p>maxn){
					y1=p+L,y2=i+L;
					maxn=i-p;
				}
				p=i;
			}
		}
		if(maxn==-1){
			printf("There are no adjacent primes.\n");
			continue;
		}
		printf("%d,%d are closest, %d,%d are most distant.\n",x1,x2,y1,y2);
	}
	return 0;
}