fzu1914

FZU 1914 Funny Positive Sequence

2011-04-10 15:35:03| 分类： FZU | 标签：fzu 1914 funny positive sequence |举报|字号 订阅

Problem 1914 Funny Positive Sequence

Time Limit: 1000 mSec Memory Limit : 32768 KB

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

Problem Description

There are n integers a1,a2,…,an-1,an in the sequence A, the sum of these n integers is larger than zero. There
are n integers b1,b2,…,bn-1,bn in the sequence B, B is the generating sequence of A and bi = a1+a2,+…+ai (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,…,an-1,an

A(1): a2,a3,…,an,a1

…

A(n-2): an-1,an,…,an-3,an-2

A(n-1): an,a1,…,an-2,an-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

Source

2010年全国大学生程序设计邀请赛（福州）

这种题目就得思维开放的思考

版本1

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
//typedef long long ll;
#define MAXN 500010
int t, n, m;
int vis[MAXN],next[MAXN],sum[MAXN],vnext[MAXN];
int a[MAXN];
int main()
{
	int i, j;
	//freopen("a.txt", "r", stdin);
	scanf("%d", &t);
	int count = 1;
	while(t--)
	{
		scanf("%d", &n);
		for(i = 0; i < n; i++) scanf("%d", &a[i]); 
		int ans = n;
		memset(vis, 0, sizeof(vis));
		memset(vnext, 0, sizeof(vnext));
		for(i = n-1; i>=0; i--)			//从后往前扫，发现-的则往前(循环)叠加至为+或者标记全满了(即ans=0)
		{
			if(a[i] <= 0 && !vis[i])
			{
				__int64 num = 0,flag=0;
				for(j = i; ans>0; j--)
				{
					if(vnext[j] && a[j]<=0)
					{
						num+=sum[j];
						j=next[j];
					}
					else num += a[j];
					if(num > 0)
					{
						sum[i]=num;
						next[i]=j;
						vnext[i]=1;
						break;
					}
					else
					{
						vis[j] = 1;
						ans--;
					}
					if(j == 0) j = n;
					else if(j == i + 1) break;
				}
			}
		}
		printf("Case %d: %d\n", count++, ans);
	}
	return 0;
}

版本2

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define MAXN 500010
int t, n, m;
int vis[MAXN],next[MAXN],sum[MAXN],vnext[MAXN];
__int64 a[MAXN];
int main()
{
	int i, j;
	scanf("%d", &t);
	int count = 1;
	while(t--)
	{
		scanf("%d", &n);
		for(i=j=0; i < n; i++) 
		{
			scanf("%I64d", &a[++j]); 
			while(a[j]<=0 && j>1 )a[j-1]+=a[j],j--;//遇到<=0的就往前压缩到为+或者j=1
		}
		n=j;
		int ans = n;
		memset(vis, 0, sizeof(vis));
		memset(vnext, 0, sizeof(vnext));
		//for(i = n; i>=1; i--)
		i=1;		//最后最多只有第一位为负数
		{
			if(a[i] <= 0 && !vis[i])
			{
				__int64 num = 0,flag=0;
				for(j = i; ans>0; j--)
				{
					if(vnext[j] && a[j]<=0)
					{
						num+=sum[j];
						j=next[j];
					}
					else num += a[j];
					if(num > 0)
					{
						sum[i]=num;
						next[i]=j;
						vnext[i]=1;
						break;
					}
					else
					{
						vis[j] = 1;
						ans--;
					}
					if(j == 1) j = n+1;
					else if(j == i + 1) break;
				}
			}
		}
		printf("Case %d: %d\n", count++, ans);
	}
	return 0;
}