FOJ 1914 Funny Positive Sequence

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Funny Positive Sequence

Accept: 422 Submit: 1860
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

There are n integers a₁,a₂,…,a_n-1,a_n in the sequence A, the sum of these n integers is larger than zero. There are n integers b₁,b₂,…,b_n-1,b_n in the sequence B, B is the generating sequence of A and bi = a₁+a₂,+…+a_i (1≤i≤n). If the elements of B are all positive, A is called as a positive sequence.

We left shift the sequence A 0,1,2,…,n-1 times, and get n sequences, that is showed as follows:

A(0): a₁,a₂,…,a_n-1,a_n

A(1): a₂,a₃,…,a_n,a₁

…

A(n-2): a_n-1,a_n,…,a_n-3,a_n-2

A(n-1): a_n,a₁,…,a_n-2,a_n-1

Your task is to find out the number of positive sequences in the set { A(0), A(1), …, A(n-2), A(n-1) }.

Input

The first line of the input contains an integer T (T <= 20), indicating the number of cases. Each case begins with a line containing one integer n (1 <= n <= 500,000), the number of elements in the sequence. The next line contains n integers ai(-2,000,000,000≤ai≤2,000,000,000,1≤i≤n), the value of elements in the sequence.

Output

For each test case, print a line containing the test case number (beginning with 1) and the number of positive sequences.

Sample Input

2
3
1 1 -1
8
1 1 1 -1 1 1 1 -1

Sample Output

Case 1: 1
Case 2: 4

题目链接：http://acm.fzu.edu.cn/problem.php?pid=1914

题目大意：给你n个数，计算前缀和，若每个前缀和都大于零，则加一；
把第一个数调到最后，计算
直至把倒数第二个数调到最后，
输出最后答案

思路分析：
其实只需要知道非正数的影响就好了。
当然；
因为有位置的调换，则可以适当的增长数组长度，将1~n-1个数在赋值到数组的后头；
从后往前扫描，为非整数则标记为1，为正数则赋值为零
最后扫描一下1~n 有多少个没有被标记则是最终答案；
代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;const int maxn=511111*2;ll num[maxn],l[maxn],r[maxn];int vis[maxn];int main(){    int T;    scanf("%d",&T);    for(int t=1;t<=T;t++)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%I64d",&num[i]);        for(int i=0;i<n-1;i++)            num[n+i]=num[i];        ll total=0;        int m=n;        n+=n-1;        int i=n-1;        memset(vis,0,sizeof(vis));        while(i>=0)        {            total+=num[i];            while(total<=0&&i>=0)            {                vis[i]=1;                i--;                total+=num[i];            }            total=0;            i--;        }        int ans=0;        for(int i=0;i<m;i++)        if(!vis[i])          ans++;        printf("Case %d: %d\n",t,ans);    }    return 0;}

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