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不好意思,本人没帮你解决问题,还给你增添了麻烦,这是我的程序(C++版)In Reply To:有一小子非说他的程序没错,要起诉北大.我觉得他挺可怜的,但我也找不出错.哪位大哥来帮看看.(C语言) Posted by:20054959 at 2007-01-16 21:09:55 不好意思,本人没帮你解决问题,还给你增添了麻烦
#include<iostream>
const long MaxXY=2000000000;
const long MaxVAB=2000000000;
const long MaxL=2100000000;
bool isMeet(long x,long y,long VA,long VB,long L,long &t); /*判断能否相见,如果能,用t返回相见时跳的次数*/
using namespace std;
void main()
{
long x,y,VA,VB,L,N;
cout<<"输入青蛙A和青蛙B位置x,y,速度VA,VB和纬线总长L,以第一个数目非法时结束"<<endl;
if(!(cin>>x))return;
while(x>=0&&x<=MaxXY)
{
if(!(cin>>y>>VA>>VB>>L))return;
if(y<0||x==y||y>MaxXY||VA<0||VA>MaxVAB||VB<0||VB>MaxVAB||L<0||L<=x||L<=y||L>MaxL)//判断输入
cout<<"错误!请重新输入!"<<endl;
else
{
if(isMeet(x,y,VA,VB,L,N))cout<<"这两只青蛙会见面的,见面时他们跳了"<<N<<"次"<<endl;
else cout<<"很遗憾,他们这样不可能见面的!"<<endl;
}
cout<<"输入青蛙A和青蛙B位置x,y,速度VA,VB和纬线总长L,以第一个数目非法时结束"<<endl;
if(!(cin>>x))return;
}
}
bool isMeet(long x,long y,long VA,long VB,long L,long &N)/*判断能否相见,如果能,用t返回相见时跳的次数*/
{
N=L;
for(int i=1;i<N;i++)
if(((VA-VB)*i+(x-y))%L==0)
{N=i;break;}
if(N!=L)return true;
return false;
}
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