编程,百马百担问题,有100匹马,驮100担货,大马驮三担,中马驮2担,两匹小马驮一担,求大、中、小各多少匹?
编程,百马百担问题,有100匹马,驮100担货,大马驮三担,中马驮2担,两匹小马驮一担,求大、中、小各多少匹?
[解决办法]
数学解方程就好了!
#include <stdio.h>
int main(void)
{
int Lar = 0, Lit = 0, Nor = 0;
for(Lit = 0; Lit < 100; Lit += 2)
{
for( Nor = 0; Nor < 100; Nor ++ )
for(Lar = 0; Lar < 100; Lar ++)
if (((Lar + Lit + Nor) == 100) && ((3 * Lar + 2 * Nor + Lit / 2) == 100))
printf("Lar = %d, Nor = %d , Lit = %d \n", Lar, Nor, Lit);
}
return 0;
}
#include <stdio.h>
int main()
{
int x, y, z;
for(z = 68; z <= 80; z += 2){
x = (z>>1)*3 - 100;
y = 200 - (z>>1)*5;
printf("大马%d匹,中马%d匹,小马%d匹\n", x, y, z);
}
return 0;
}
#include <iostream>
using namespace std;
int main()
{
for(int n1=0;n1<100;n1++){
for(int n2=0;n2<100-n1;n2++){
int n3=100-n1-n2;
if(3*n1+2*n2+n3/2==100 && n3%2==0)//n3可为奇数的话,那么有一匹小马是多余的
cout<<"Lar = "<<n1<<", Nor = "<<n2<<" , Lit = "<<n3<<endl;
}
}
return 0;
}