大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升 (正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。
Input三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。Output如果能平分的话请输出最少要倒的次数，否则输出"NO"。Sample Input
7 4 34 1 30 0 0
Sample Output
NO3
#include#include #include #include #include #include using namespace std;typedef long long LL;#define Mem0(x) memset(x, 0, sizeof(x))#define MemI(x) memset(x, -1, sizeof(x))#define MemM(x) memset(x, 0x3f, sizeof(x))const int MAXN = 10005;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;//vis -> all of case of n and mint vis[150][150], s, n, m;struct Node{ int s, n, m, cnt;};queue q;int bfs(){ while(!q.empty()) q.pop(); Mem0(vis); Node now, next; now.s = s, now.n = 0, now.m = 0, now.cnt = 0; q.push(now); vis[now.n][now.m] = 1; while(!q.empty()) { now = q.front(); q.pop();// cout << now.s << " " << now.n << " " << now.m << " " << now.cnt << endl; int cnt = 0; if(2 * now.s == s) cnt++; if(2 * now.n == s) cnt++; if(2 * now.m == s) cnt++; if(cnt == 2) return now.cnt; // s -> n next.cnt = now.cnt + 1; if(now.s && now.n != n) { int d = n - now.n; next.s = max(0, now.s - d); next.n = min(n, now.n + now.s); next.m = now.m; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } // s -> m if(now.s && now.m != m) { int d = m - now.m; next.s = max(0, now.s - d); next.m = min(m, now.m + now.s); next.n = now.n; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } // n -> s if(now.n && now.s != s) { int d = s - now.s; next.n = max(0, now.n - d); next.s = min(s, now.s + now.n); next.m = now.m; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } // n -> m if(now.n && now.m != m) { int d = m - now.m; next.n = max(0, now.n - d); next.m = min(m, now.m + now.n); next.s = now.s; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } //m -> s if(now.m && now.s != s) { int d = s - now.s; next.m = max(0, now.m - d); next.s = min(s, now.s + now.m); next.n = now.n; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } // m -> n if(now.m && now.n != n) { int d = n - now.n; next.m = max(0, now.m - d); next.n = min(n, now.n + now.m); next.s = now.s; if(!vis[next.n][next.m]) { vis[next.n][next.m] = 1; q.push(next); } } } return 0;}int main(){ while(cin >> s >> n >> m) { if(!s && !n && !m) break; if(s % 2) { cout << "NO" << endl; continue; } else { int ans = bfs(); if(ans) cout << ans << endl; else cout << "NO" << endl; } } return 0;}