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:
    [转]红帽 Red Hat Linux相关产品iso镜像下载【百度云】
    JAVA中的类
    Java并发编程:Lock
    字符集和编码的区别
    MySQL索引背后的数据结构及算法原理
    B树、B-树、B+树、B*树 红黑树
    linux下nginx的安装
    对.net orm工具Dapper在多数据库方面的优化
    Dapper使用方法
    filebeat to elasticsearch配置
  • 原文地址:https://www.cnblogs.com/yuhan-blog/p/12308976.html
Copyright © 2011-2022 走看看