1217: VIJOS-P1042
时间限制: 0 Sec 内存限制: 128 MB
提交: 78 解决: 29
[提交][状态][讨论版]
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。
=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;
}