东大OJ-5到100000000之间的回文质数

1217: VIJOS-P1042

时间限制: 0 Sec  内存限制: 128 MB

提交: 78  解决: 29

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题目描述

        有一天，雄霸传授本人风神腿法第一式：捕风捉影..............的步法（弟子一：堂主，你大喘气呀。风：你给我闭嘴。）捕风捉影的关键是换气（换不好就会大喘气...）。         使用捕风捉影这一招时并不是每一步都喘气，而是在特定的步数喘气。一般来说功力越高，喘气越稀疏。喘气的步数符合特定规律：第一要是SUSHU（弟子二：哇塞！堂主，你还会鸟语，我好好崇拜你呦！可是SUSHU是什么意思呢？风：笨蛋，那是汉语拼音！）第二要是一个回文数，回文数就是正反念一样的数，如：123321，121，5211314（弟子三：堂主，最后一个好象不是...风：废话，当然不是了，我是考察一下你们的纠错能力！）现在给出两个数M，N(5<
=M< N< =100,000,000)，你要算出M，N之间需要换气的都有哪几步。（包括M，N）。算出来的可以提升为本堂一级弟子，月薪（1000000000000000000000000000000000000000000  MOD  10  ）元。

输入

两个整数M，N。用空格隔开。

输出

在M，N之间的换气点，每个一行。

样例输入

100 500

样例输出

101
131
151
181
191
313
353
373
383

#include<stdio.h>
#include<string.h>
int prime[4000];
int psize;
int from, to;
void getPrime(){
	memset(prime, 0, sizeof(prime));
	psize = 0;
	int i,j;
	bool a[10000];
	memset(a, -1, sizeof(a));
	for (i = 2; i < 10000; i++){
		if (!a[i])continue;
		prime[psize++] = i;
		for (j = i + i; j < 10000; j+=i)a[j] = false;
	}
}
bool isPrime(int n){
	int i;
	for (i = 0;prime[i]*prime[i]<=n; i++)if (n%prime[i] == 0)return false;
	return true;
}
void one(){
	if (from>7)return;
	if (from == 5)printf("5\n");
	printf("7\n");
}
void three(){
	if (from > 999)return;
	int i, j,k;
	int a[4] = { 1, 3, 7, 9 };
	for (i = 0; i < 4;i++)
	for (j = 0; j < 10; j++){
		k = a[i] * 100 + j * 10 + a[i];
		if ( k< from)continue;
		if (k>to)return;
		if (isPrime(k))printf("%d%d%d\n", a[i], j,a[ i]);
	}
}
void five(){
	if (from>99999)return;
	int a[4] = { 1, 3, 7, 9 };
	int i, j, k, l;
	for (i = 0; i < 4;i++)
	for (j = 0; j < 10;j++)
	for (k = 0; k < 10; k++){
		l = a[i] * 10000 + j * 1000 + k * 100 + j * 10 + a[i];
		if (l<from)continue;
		if (l>to)return;
		if (isPrime(l))printf("%d%d%d%d%d\n", a[i], j, k, j, a[i] );
	}
}
void seven(){
	if (from>9999999)return;
	int a[4] = { 1, 3, 7, 9 };
	int i, j, k, l,m;
	for (i = 0; i < 4; i++)
	for (j = 0; j < 10; j++)
	for (k = 0; k < 10; k++)
	for(m=0;m<10;m++){
		l = a[i] * 1000000 + j * 100000 + k * 10000 + m * 1000 +k*100+j*10+ a[i];
		if (l<from)continue;
		if (l>to)return;
		if (isPrime(l))printf("%d%d%d%d%d%d%d\n", a[i], j, k,m,k, j, a[i]);
	}
}
int main(){
	getPrime();
	prime[psize++]=10001;
	scanf("%d%d", &from, &to);
	one();
	if (11 >= from && 11 <= to)printf("11\n");
	three(); five(); seven();
	return 0;
}


上面这个5ms

再慢一点的算法:6311ms

#include<stdio.h>
#include<string.h>
#include<math.h>
int prime[4000];
int psize;
int from, to;
void getPrime(){
	memset(prime, 0, sizeof(prime));
	psize = 0;
	int i,j;
	bool a[10000];
	memset(a, -1, sizeof(a));
	for (i = 2; i < 10000; i++){
		if (!a[i])continue;
		prime[psize++] = i;
		for (j = i + i; j < 10000; j+=i)a[j] = false;
	}
}
bool isPrime(int n){
	int i;
	for (i = 0;prime[i]*prime[i]<=n; i++)if (n%prime[i] == 0)return false;
	return true;
}
bool isHuiwen(int n,int wei){
	if (wei == 0||wei==1)return true;
	int i;
	if (n /(int) pow((double)10, wei-1) == n % 10){
		n %= (int)pow((double)10, wei-1);
		n /= 10;
		if (isHuiwen(n, wei - 2))return true;
	}
	return false;
}
int main(){
	getPrime();
	prime[psize++]=10001;
	scanf("%d%d", &from, &to);
	if (from == 5)printf("5\n");
	if (from <= 7 && to >= 7)printf("7\n");
	if (11 >= from && 11 <= to)printf("11\n");
	if (from < 100)from = 101;
	for (; from <= to; from++){
		int wei = log10((double)from)+1;
		if (wei % 2 == 0){
			from = pow((double)10, wei);
			wei++;
			if (from>to)break;
		}
		if (isPrime(from) && isHuiwen(from,wei))
			printf("%d\n", from);
	}
	return 0;
}
最慢42805ms

#include<stdio.h>
#include<string.h>
#include<math.h>
int prime[4000];
int psize;
int from, to;
void getPrime(){
	memset(prime, 0, sizeof(prime));
	psize = 0;
	int i,j;
	bool a[10000];
	memset(a, -1, sizeof(a));
	for (i = 2; i < 10000; i++){
		if (!a[i])continue;
		prime[psize++] = i;
		for (j = i + i; j < 10000; j+=i)a[j] = false;
	}
}
bool isPrime(int n){
	int i;
	for (i = 0;prime[i]*prime[i]<=n; i++)if (n%prime[i] == 0)return false;
	return true;
}
bool isHuiwen(int n,int wei){
	if (wei == 0||wei==1)return true;
	int i;
	if (n /(int) pow((double)10, wei-1) == n % 10){
		n %= (int)pow((double)10, wei-1);
		n /= 10;
		if (isHuiwen(n, wei - 2))return true;
	}
	return false;
}
int main(){
	getPrime();
	prime[psize++]=10001;
	scanf("%d%d", &from, &to);
	if (from == 5)printf("5\n");
	if (from <= 7 && to >= 7)printf("7\n");
	if (11 >= from && 11 <= to)printf("11\n");
	if (from < 100)from = 101;
	for (; from <= to; from++){
		if (isPrime(from) && isHuiwen(from,log10((double)from)+1))
			printf("%d\n", from);
	}
	return 0;
}