poj3061 Subsequence

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//O(nlogn)#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX_N = 1e5;int n, S;int a[MAX_N], sum[MAX_N + 1];void solve(){memset(sum, 0, sizeof(sum));// 计算 sumfor (int i = 0; i < n; i++)sum[i + 1] = sum[i] + a[i];if (sum[n] < S){cout << 0 << endl;return;} int res = n;for (int s = 0; sum[s] + S <= sum[n]; s++){// 利用二分搜索法求出tint t = lower_bound(sum + s, sum + n, sum[s] + S) - sum - s; res = min(res, t);}cout << res << endl;}int main(){int t;cin >> t;while (t--){cin >> n >> S;for (int i = 0; i < n; i++) cin >> a[i];solve(); }return 0;}

//优化：O(n)#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX_N = 1e5;int n, S;int a[MAX_N], sum[MAX_N + 1];void solve(){int res = n + 1;int s = 0, t = 0, sum = 0;  // 初始化for (; ; ){while (t < n && sum < S) // 只要 t没有超出范围， 且sum没达到S，就不断将sum增加a[t]，并且将t增加1 {sum += a[t++];}if (sum < S) break; // 若所有数加起来，仍小于S，则此后不可能再有res的更新，可以退出循环，否则更新res res = min(res, t - s);sum -= a[s++]; //  将sum减去as，s增加1后，回到s不断增加at，t++的循环 }if (res > n) res = 0;cout << res << endl;}int main(){int t;cin >> t;while (t--){cin >> n >> S;for (int i = 0; i < n; i++) cin >> a[i];solve(); }return 0;}

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