2 条题解
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0
#include<iostream> using namespace std; const int maxn = 10000; int n; int a[maxn]; int b[maxn]; int f(int l, int r, int depth) { if (l > r) return 0; int min = maxn, mink; for(int i = 1;i <= r;++i){ if (min > a[i]) { min = a[i]; mink = i; } } int lres = f(l, mink - 1, depth + 1); int rres = f(mink + 1, r, depth + 1); return lres + rres + depth \* b[mink]; } int main() { cin >> n; for(int i = 0;i < n;++i) cin >> a[i]; for(int i = 0;i < n;++i) cin >> b[i]; cout << f(0, n - 1, 1) << endl; return 0; } -
0
#include using namespace std; const int maxn = 10000; int n; int a[maxn]; int b[maxn]; int f(int l, int r, int depth) { if (l > r) return 0; int min = maxn, mink; for(int i = 1;i <= r;++i){ if (min > a[i]) { min = a[i]; mink = i; } } int lres = f(l, mink - 1, depth + 1); int rres = f(mink + 1, r, depth + 1); return lres + rres + depth * b[mink]; } int main() { cin >> n; for(int i = 0;i < n;++i) cin >> a[i]; for(int i = 0;i < n;++i) cin >> b[i]; cout << f(0, n - 1, 1) << endl; return 0; }
- 1
信息
- ID
- 15
- 时间
- 1000ms
- 内存
- 64MiB
- 难度
- 7
- 标签
- 递交数
- 32
- 已通过
- 8
- 上传者