Non-negative Partial Sums HDU

题目链接:点我

---

    You are given a sequence of n numbers a 0,..., a n-1. A cyclic shift by k positions (0<=k<=n-1) results in the following sequence: a k a k+1,..., a n-1, a 0, a 1,..., a k-1. How many of the n cyclic shifts satisfy the condition that the sum of the fi rst i numbers is greater than or equal to zero for all i with 1<=i<=n?

Input

    Each test case consists of two lines. The fi rst contains the number n (1<=n<=10 6), the number of integers in the sequence. The second contains n integers a 0,..., a n-1 (-1000<=a i<=1000) representing the sequence of numbers. The input will finish with a line containing 0.

Output

    For each test case, print one line with the number of cyclic shifts of the given sequence which satisfy the condition stated above.

Sample Input

32 2 13-1 1 11-10

Sample Output

320

---

题意:

给你一个循环序列,求序列中满足每一项的前缀和都大于等于0的序列的个数.

思路:

序列倍增然后求前缀和,然后维护一个前缀和单调递增的单调队列,

代码:

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>using namespace std;const int maxn = 2e6 + 10;int sum[maxn];int a[maxn];int main(){    int n;    while(scanf("%d", &n), n){        for(int  i = 1;i <= n; ++i){            scanf("%d", &sum[i]);            sum[n+i] = sum[i];            sum[i] += sum[i-1];        }int ans = 0;        for(int i = n+1; i < n+n; ++i){            sum[i] = sum[i-1] + sum[i];        }int l =1,r = 0;        for(int  i = 1; i < n+n; ++i){            while(r>=l && i - a[l]-1>= n)                ++l;            while(r >= l && sum[a[r]] >= sum[i])                --r;            a[++r] = i;            if( i >= n && sum[a[l]] - sum[i-n] >=0)                ++ans;        }        printf("%d\n",ans);    }    return 0;}