求助
求大佬们给咱解析解析下面的代码，希望能详细一点，谢谢各位大佬。#include <algorithm>
#include <cstdio>
using namespace std;
int s = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int main()
{
do
{
if(!s || !s || !s) continue;
int a = s * 1000 + s * 100 + s * 10 + s;
int b = s * 1000 + s * 100 + s * 10 + s;
int c = s * 10 + s;
if((a - b) * c == 900) printf("(%d-%d)*%d=900\n", a, b, c);
}while(next_permutation(s, s + 10));
return 0;
} #include <algorithm>
#include <cstdio>
using namespace std;
int s = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int main()
{
do
{
if(!s || !s || !s) continue; //当这几个索引位置的值为0时跳过本次循环
int a = s * 1000 + s * 100 + s * 10 + s; //将数组的前四个整数组成一个四位数a
int b = s * 1000 + s * 100 + s * 10 + s;//后续四个整数组成另一个四位数b
int c = s * 10 + s; //最后两个整数组成一个两位数c
if((a - b) * c == 900) printf("(%d-%d)*%d=900\n", a, b, c); //当(a - b) * c == 900 输出
}while(next_permutation(s, s + 10));
return 0;
}
next_permutation(s, s + 10)这里讲一下next_permutation()函数其实就是全排列函数
给个例子自己看，或者自己查查
vector<char> chars = {'a', 'b', 'c'};
do {
cout << chars << chars << chars << endl;
} while (next_permutation(chars.begin(), chars.end()));
输出为：
abc
acb
bac
bca
cab
cba
谢谢大佬
页:
[1]