给你一堆物品,每个物品只能用一次,求物品能组合的所有情况
传送门
用01背包,初始化dp[0] = 1
然后01背包或取操作即可,时间复杂度(O(mn))
int a[N], dp[N];
void solve(int kase){
int n = read(), sum = 0;
for(int i = 1; i <= n; i++) a[i] = read(), sum += a[i];
dp[0] = 1;
for(int i = 1; i <= n; i++) { // 求出放一边的情况
for(int j = sum; j >= a[i]; j--) {
dp[j] |= dp[j - a[i]];
}
}
for(int i = 1; i <= n; i++) { // 两边都可以放的情况
for(int j = 1; j <= sum - a[i]; j++) {
dp[j] |= dp[j + a[i]];
}
}
int k = read();
for(int i = 1; i <= k; i++) {
int x = read();
printf("%s
", x <= sum && dp[x] ? "YES" : "NO");
}
}