201909-2 小明种苹果（续）C++（100分）

<1000 的输入输出使用cin cout，否则使用C语言的标准输入输出

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 1000;
int at[N + 10];  // array temp

int main() {
    
    
//	freopen("in.txt", "r", stdin);
	memset(at, 0, sizeof(at));
	int sum = 0;
	int subsum = 0;
	vector<int> s;

	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
    
    
		s.push_back(0);

		int m;
		scanf("%d", &m);
		scanf("%d", &subsum);
		for (int j = 2; j <= m; j++) {
    
    
			int t;
			scanf("%d", &t);

			if (t <= 0) {
    
    
				subsum += t;
			} else if (subsum > t) {
    
    
				s[i - 1] = 1;
				subsum = t;
			}
		}
		sum += subsum;
	}
//	for (int i = 0; i < s.size(); i++) {
    
    
//		cout << s[i] << endl;
//	}
	int cnt = count(s.begin(), s.end(), 1);
	for (int i = 0; i < s.size(); i++) {
    
    
		at[i] = s[i] && s[(i + 1) % s.size()] && s[(i + 2) % s.size()];
	}
	int con_num = count(at, at + s.size(), 1);  // continuous number 3 times
	printf("%d %d %d\n", sum, cnt, con_num);

	return 0;
}

/**
in.txt

4
4 74 -7 -12 -5
5 73 -8 -6 59 -4
5 76 -5 -10 60 -2
5 80 -6 -15 59 0

5
4 10 0 9 0
4 10 -2 7 0
2 10 0
4 10 -3 5 0
4 10 -1 8 0

222 1 0

39 4 2
*/