按题意，比较到站的最大值.

using ll = long long;
int a[105], b[105], c[105];
int main() {
    // ios_base::sync_with_stdio(false), cin.tie(0);
    int _;
    for (cin >> _; _--;) {
        int n, s = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &c[i]), s += a[i] - b[i - 1] + c[i];
            if (i == n) break;
            s = max(s + (b[i] - a[i] + 1) / 2, b[i]);
        }
        cout << s << "\n";
    }
    return 0;
}

Problem 1501B. Napoleon Cake

从后往前比较.

int arr[200001];
int res[200001];
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int _ = 1;
    for (cin >> _; _--;) {
        int n, i, mn = 1e9;
        cin >> n;
        for (i = 1; i <= n; i++) cin >> arr[i];
        for (i = n; i >= 1; i--) {
            mn = min(mn, i - arr[i]);
            res[i] = (mn < i);
        }
        for (i = 1; i <= n; i++) cout << res[i] << " ";
        cout << endl;
    }
    return 0;
}

Problem 1501C. Going Home

把两数之和想象成坐标，x 和 y

使用双指针然后存储和存储一下，如果能匹配并且坐标不重复即可输出。

数组开大点，这个导致WA好几次了....

const int N = 5000003;
int x[N], y[N], a[N];
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j) {
            int s = a[i] + a[j];
            if (x[s] && y[s] && x[s] != i && y[s] != j && x[s] != j &&
                y[s] != i) {
                cout << "YES\n";
                cout << i << " " << j << " " << x[s] << " " << y[s] << "\n";
                return 0;
            }
            x[s] = i;
            y[s] = j;
        }
    cout << "NO\n";
    return 0;
}