zoukankan      html  css  js  c++  java
  • A. Drazil and Date

    Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y)(x - 1, y)(x, y + 1) or (x, y - 1).

    Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.

    Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?

    Input

    You are given three integers ab, and s ( - 109 ≤ a, b ≤ 1091 ≤ s ≤ 2·109) in a single line.

    Output

    If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).

    Otherwise, print "Yes".

    Sample test(s)
    input
    5 5 11
    
    output
    No
    
    input
    10 15 25
    
    output
    Yes
    
    input
    0 5 1
    
    output
    No
    
    input
    0 0 2
    
    output
    Yes
    这题要注意数据很大,要用__int64,而且要注意负数,0,0的时候要特殊考虑。
    #include<stdio.h>
    #include<math.h>
    int main()
    {
    	__int64 n,m,i,j,k,d;
    	while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
    	{
    		if(n==0 && m==0){
    			if(k%2==0){
    				printf("Yes
    ");continue;
    			}
    			else{
    				printf("No
    ");continue;
    			}
    		}
    		d=fabs(n)+fabs(m);
    		if(d>k){
    			printf("No
    ");continue;
    		}
    		if(d==k){
    			printf("Yes
    ");continue;
    		}
    		if(d<k){
    			if((k-(d-1))%2==1){
    				printf("Yes
    ");
    			}
    			else printf("No
    ");
    		}
    	}
    }
    
  • 相关阅读:
    基于php socket(fsockopen)的应用实例分析
    php多线程的问题
    PHP之fsockopen提交POST数据讲解
    Sort List
    Insertion Sort List
    LRU Cache
    Reorder List
    Word Break II
    Word Break
    Copy List with Random Pointer
  • 原文地址:https://www.cnblogs.com/herumw/p/9464849.html
Copyright © 2011-2022 走看看