HDU 5783 Divide the Sequence（数列划分）
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description - 题目描述
Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not small than 0.
Alice有一个数列A，她想把A拆分出尽可能多的子串，并且每个子串的任意前缀和都大于0。
CN
Input - 输入
The input consists of multiple test cases.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤1e6
−10000≤A[i]≤10000
You can assume that there is at least one solution.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤1e6
−10000≤A[i]≤10000
You can assume that there is at least one solution.
多组测试用例。
每组测试用例的第一行为一个整数n。
下一行有n个整数A1,A2⋯An。
≤n≤1e6
−≤A[i]≤
你可以认为不存在无解的情况。
CN
Output - 输出
For each test case, output an integer indicates the maximum number of sequence division.
对于每组测试用例，输出此序列的最大划分数。
CN
Sample Input - 输入样例
6
1 2 3 4 5 6
4
1 2 -3 0
5
0 0 0 0 0
Sample Output - 输出样例
6
2
5
题解
贪心，一开始看错题目了，一直想不出解法，浪费了巨多时间……
某子串1 2 3的所有前缀为
1
1 2
1 2 3
显然如果出现负数的话，需要前面的正数来凑，反正无法满足条件。
做法就是从后往前扫描，记录临时的后n项和，>=0则清空，划分数+1。
极限数据应该是5e5 * 10000 = 5e9，单纯的int会超，需要__int64。
代码 C++
#include <cstdio>
__int64 data[];
int main(){
__int64 n, i, opt, tmp;
while (~scanf("%I64d", &n)){
for (i = opt = tmp = ; i < n; ++i) scanf("%I64d", data + i);
for (--i; ~i; --i){
tmp += data[i];
if (tmp >= ) tmp = , ++opt;
}
printf("%I64d\n", opt);
}
return ;
}