題意 ：

• 給定長度為n的序列和數字s，找出最長的區間$[l,r]$，使得$?i\in [l,r],pr{e}_i?pr{e}_{l?1}+s\ge 0$，其中$pr{e}_i$表示前綴和

思路 ：

• 考慮雙指針，對某一段$[l,r]$區間滿足答案要求，則更新答案，且右指針移動
• 若右指針移動后不滿足條件，則不斷移動左指針直到滿足條件
• 注意如何設置初始值，來應對無解的情況

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <unordered_set> #include <math.h> using namespace std;typedef long long ll; //typedef pair<int, int> PII;#define endl '\n' #define fi first #define se second #define push_back #define rep(i, l, r) for (ll i = l; i <= r; i ++ )#define int long longconst int N = 2e5 + 10;int a[N];void solve() {int n, s; cin >> n >> s;for (int i = 1; i <= n && cin >> a[i]; i ++ ) a[i] += a[i - 1];int l = 1, r = 0;for (int j = 1, i = 1; j <= n; j ++ ){while (i <= j && a[j] - a[i - 1] + s < 0) i ++ ;if (r - l + 1 < j - i + 1)l = i, r = j;}if (r == 0) cout << -1 << endl;else cout << l << ' ' << r << endl; }signed main() {cin.tie(nullptr) -> sync_with_stdio(false);int _;cin >> _;while (_ -- )solve();// return 0; } #include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <unordered_set> #include <math.h> using namespace std;//typedef long long ll; typedef pair<int, int> PII;#define endl '\n' #define fi first #define se second #define push_back #define rep(i, l, r) for (ll i = l; i <= r; i ++ )#define int long longconst int N = 2e5 + 10;int n, s; int a[N]; int l, r, ansl, ansr, sum;void init() {l = r = 1, sum = 0, ansl = ansr = -1; }void update() {if (r - l > ansr - ansl) ansl = l, ansr = r; }void solve() {cin >> n >> s;for (int i = 1; i <= n && cin >> a[i]; i ++ );init();while (l <= n && r <= n + 1){if (sum + s >= 0) update(), sum += a[r ++ ];else sum -= a[l ++ ];}if (ansr == -1) cout << -1 << endl;else cout << ansl << ' ' << ansr - 1 << endl; }signed main() {cin.tie(nullptr) -> sync_with_stdio(false);int _;cin >> _;while (_ -- )solve();// return 0; }

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