zoukankan      html  css  js  c++  java
  • Codeforces 231A

    题目: A. Team

    time limit per test: 2 seconds
    memory limit per test: 256 megabytes
    input: standard input
    output: standard output

    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.

    思路:

    1.每行相加大于1记一次数即可;

    代码:

    #include<iostream>
    using namespace std;
    int main(){
    	int n,cnt=0;
    	scanf("%d",&n);
    	for(int i=0;i<n;i++){
    		int a,b,c,t;
    		scanf("%d%d%d",&a,&b,&c);
    		t=a+b+c;
    		if(t>1) cnt++;
    	}
    	printf("%d",cnt);
    	return 0;
    }
    
  • 相关阅读:
    faster rcnn流程
    本质矩阵E求解及运动状态恢复
    对极几何
    仿射变换与投影变换
    SVD分解及线性最小二乘问题
    常见的网络结构
    Docker swarm 获取service的container信息
    XNginx升级记录
    sparql 查询语句快速入门
    开源音乐下载神器XMusicDownloader更新,支持歌单一键下载,支持无损音乐
  • 原文地址:https://www.cnblogs.com/yuhan-blog/p/12308975.html
Copyright © 2011-2022 走看看