1 条题解
-
0
参考答案:
#include<iostream> using namespace std; const int N = 10010, INF = 1e9; int q[N]; int res; int main() { int n; cin >> n; for (int i = 0; i < n; ++i) scanf("%d", &q[i]); for (int i = 0; i < n; ++i) { int minn = INF, maxx = -INF; for (int j = i; j < n; ++j) { minn = min(minn, q[j]); maxx = max(maxx, q[j]); if ((maxx - minn) == j - i) res++; } } cout << res << endl; return 0; }
- 1
信息
- ID
- 5384
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 1
- 标签
- 递交数
- 15
- 已通过
- 5
- 上传者