[CF从零单排#2]CF231A - Team

题目来源：http://codeforces.com/problemset/problem/231/A

One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.

This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.

Input
The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.

Output
Print a single integer — the number of problems the friends will implement on the contest.

Examples
input
3
1 1 0
1 1 1
1 0 0
output
2
input
2
1 0 0
0 1 1
output
1
Note
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.

In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.

题目大意：

三个朋友组队打编程比赛，当有两个以上的人有把握解决问题的时候，就认为可以解决这题。第一行输入问题数量n, 接下来的每一行，都有三个数字，表示三个人是否有把握解决问题，有把握用1，没把握用0。问最后有把握做多少题。

题目分析：

显然，只需要将每一行的三个数加起来，用sum存起来，如果sum>=2说明有把握解决，标记数量加1。

参考代码：

#include <bits/stdc++.h>
using namespace std;
int main(){
	int n, a, b, c, ans = 0;
	cin >> n;
	for(int i=1; i<=n; i++){
		cin >> a >> b >> c;
		a += b + c;
		if(a>=2)
			ans ++; 
	}
	cout << ans;
	return 0;
}