POJ1742 Posted on 2017-01-18 | In Competitive Programming , POJ CoinsSolution sketch有限背包問題,但是正常做會TLE QQ!(原理請看解說) AC Code12345678910111213141516171819202122232425262728293031323334353637383940#include <cstdio>#include <cstring>bool ok[111111];int used[111111];int main(){ int n, m; while (scanf("%d %d", &n, &m) == 2 && (n || m)) { int val[n], cnt[n]; for (int i = 0; i < n; i++) scanf("%d", &val[i]); for (int i = 0; i < n; i++) scanf("%d", &cnt[i]); memset(ok, 0, sizeof(ok)); ok[0] = true; for (int i = 0; i < n; i++) { // 對於所有的硬幣種類一個一個看 memset(used, 0, sizeof(used)); for (int j = 0; j <= m - val[i]; j++) { // 上限訂定為至少用一個硬幣的價值 int next = j + val[i]; // 如果目前這個數字已經可以湊出來,且還有剩餘的第i種硬幣可用,且利用一個第i種硬幣的那個值還未組出 if (ok[j] == true && cnt[i] - used[j] > 0 && next <= m && ok[next] == false) { ok[next] = true; used[next] = used[j] + 1; // 用一個硬幣掉 } } } int ans = 0; for (int i = 1; i <= m; i++) { if (ok[i] == true) ans++; } printf("%d\n", ans); } return 0;}