NYOJ488 素数环 【回溯】+【预处理】

原题链接

这题要先对n进行奇偶数判断，如果是奇数且不为1则NO answer，否则有解，回溯加剪枝解决。

#include <stdio.h>int n, store[21], Prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 };bool hasSelect[21], ok;bool checkPrime(int k){for(int i = 0; i < 14; ++i)if(k == Prime[i]) return 1;return 0;}void backTrack(int k){if(k == n + 1){if(checkPrime(store[1] + store[n])){printf("%d", store[1]);for(int i = 2; i <= n; ++i) printf(" %d", store[i]);printf("\n");ok = 1;}return;}for(int i = 2; i <= n; ++i){if(!hasSelect[i] && checkPrime(i + store[k - 1])){hasSelect[i] = 1;store[k] = i;backTrack(k + 1);hasSelect[i] = 0;}}}int main(){int i = 1;while(scanf("%d", &n), n){printf("Case %d:\n", i++);if(n & 1){if(n == 1) printf("1\n");else printf("No Answer\n");continue;}ok = 0;store[1] = 1;backTrack(2);if(!ok) printf("No Answer\n");}return 0;}

815818长木素数环Accepted28232C/C++04-15 10:46:05