解题思路:
int er(int a)
{
int sum1=0,sum0=0,b=a;
for(;;)
{
if(b%2==1)
{
sum1++;
b=(b-1)/2;
}
else if(b%2==0)
{
sum0++;
b=b/2;
} //判断1,0的个数;
if(b==0)//B无意义后终止循环返回各值;
{
sum1;//本数二进制中1的个数;
sum0;//本数二进制中0的个数;
}
}
}
注意事项:
以上函数需写两遍以便返回两个不同的值
参考代码:
#include <bits/stdc++.h> using namespace std; int er1(int a) { int sum1=0,sum0=0,b=a; for(;;) { if(b%2==1) { sum1++; b=(b-1)/2; } else if(b%2==0) { sum0++; b=b/2; } if(b==0) { return sum1; } } } int er0(int a) { int sum1=0,sum0=0,b=a; for(;;) { if(b%2==1) { sum1++; b=(b-1)/2; } else if(b%2==0) { sum0++; b=b/2; } if(b==0) { sum0-1; return sum0; } } } int main() { int sum=0,sum1=0; for(int i=1; i<=1000; i++) { if(er1(i)>er0(i)) { sum=sum+1; } if(er1(i)<=er0(i)) { sum1=sum1+1; } } cout<<sum<<" "<<sum1; return 0; }
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