So, for example, if one player claims 343 points and the other claims 49, then clearly the first player is lying; the only way to score 343 is by crashing balloons labeled 7 and 49, and the only way to score 49 is by crashing a balloon labeled 49. Since each of two scores requires crashing the balloon labeled 49, the one claiming 343 points is presumed to be lying.
On the other hand, if one player claims 162 points and the other claims 81, it is possible for both to be telling the truth (e.g. one crashes balloons 2, 3 and 27, while the other crashes balloon 81), so the challenge would not be upheld.
By the way, if the challenger made a mistake on calculating his/her score, then the challenge would not be upheld. For example, if one player claims 10001 points and the other claims 10003, then clearly none of them are telling the truth. In this case, the challenge would not be upheld.
Unfortunately, anyone who is willing to referee a game of crashing balloon is likely to get over-excited in the hot atmosphere that he\she could not reasonably be expected to perform the intricate calculations that refereeing requires. Hence the need for you, sober programmer, to provide a software solution.
Input
Pairs of unequal, positive numbers, with each pair on a single line, that are claimed scores from a game of crashing balloon.
Output
Numbers, one to a line, that are the winning scores, assuming that the player with the lower score always challenges the outcome.
Sample Input
343 49 3599 610 62 36
Sample Output
49 610 62
这道题的难点在于两位选手都可能有多个不同的分解方案,只要其中任何一种分解方案不冲突即是被挑战者赢。参考代码如下:
#include<stdio.h> bool aTrue, bTrue; int judge(int m, int n, int p) { if(aTrue) return 0; if(m == 1 && n == 1) { aTrue = true; return 0; } if(n == 1) bTrue = true; while(p > 1) { if(m%p == 0) judge(m/p, n, p-1); if(n%p == 0) judge(m, n/p, p-1); p--; } return 0; } int main() { int a, b; while(scanf("%d%d", &a, &b)!=EOF) { if(a < b) {int temp = a; a = b; b = temp;} aTrue = false; bTrue = false; judge(a, b, 100); if(!aTrue && bTrue) printf("%d\n",b); else printf("%d\n",a); } return 0; }
#include <stdio.h>
int flag1, flag2; //分别表示n, m是否已被成功分解 void dfs(int n, int m, int fac = 100){ if(flag1) //如果n, m均可分解且因子不同 return; if(n == 1 && m == 1){ //n, m均分解完 flag1 = 1; flag2 = 1; return; } if(m == 1){ //m分解完 flag2 = 1; } if (fac < 2) return; if(n % fac == 0) //精华之处,对同一个因子不同时分解n, m,难点解决 dfs(n / fac, m, fac - 1); if(m % fac == 0) dfs(n, m / fac, fac - 1); dfs(n, m, fac - 1); } int main(){ int n, m; while(~scanf("%d%d", &n, &m)){ if(m > n){ //交换m, n n = m ^ n; m = m ^ n; n = m ^ n; } flag1 = 0; flag2 = 0; dfs(n, m); if(flag1 || !flag2) //如果n成功分解货m无法成功分解 printf("%d\n", n); else printf("%d\n", m); } return 0; }
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