九度 oj 题目1550：分糖果

http://ac.jobdu.com/problem.php?pid=1550

默认给最少“1”， 如果a[i] > a[i-1], 比前面多给一个。如果 a[i] < a[i-1], a[i] 给“1”个，检查a[i]前面的序列 如果不满足条件的就多给糖果

参考了http://blog.csdn.net/xiexingshishu/article/details/20319825

#include <cstdio>#include <algorithm>using namespace std;#define rep(i,j,k) for(int i=j;i<=k;i++)#define per(i,j,k) for(int i=k;i>=j;i--)#define inone(i) scanf("%d",&i)const int maxn = 1e5 + 10;int a[maxn], n, r, cnt, t[maxn];int main(){while (inone(n) != EOF){cnt = 0;rep(i, 0, n - 1){inone(a[i]);t[i] = 1;if (i == 0) continue;if (a[i - 1] < a[i]) t[i] = t[i - 1] + 1;else if (a[i - 1] > a[i]){for (int j = i - 1; j >= 0 && a[j] > a[j + 1]; j--)t[j] = max(t[j],t[j + 1] + 1);}}rep(i, 0, n - 1) cnt += t[i];printf("%d\n",cnt);}    return 0;}