枚舉

列出所有可能

聰明的枚舉

/*
 * NTUCPC
 * 湊平方數
*/
#include <bits/stdc++.h>
using namespace std;

int main() {
    int N;
    cin >> N;
    for(int c=sqrt(N);c>=1;--c){
        for(int a=1;a<c*c;++a){
            int b=c*c-a;
            cout << "[" << a << "," << b << "," << c << "]" << endl;
        }
    }
}

子集合

/*
 * AP325
 * P-1-7. 子集合乘積
*/
#include<bits/stdc++.h>
using namespace std;
int n;

int f(int a[],int i,int sum){
	if(i>=n){
		cout << sum << endl;
		return 0;
	}
	f(a, i + 1, ((sum==0) ? 1 : sum) * a[i]);
    f(a, i + 1, sum);
}

int main(){
	cin >> n;
	int a[n];
	int p=10009;
	for(int i=0;i<n;i++){
		cin >> a[i];
	}
	f(a,0,0);
}

next_permutatio([ ],[ ]+n)

#include <bits/stdc++.h>
using namespace std;
 
int a[6][6];
int cnt[6];
 
int main() {
    
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 6; j++){
            cin >> a[i][j];
        }
        cin >> cnt[i];
    }
 
    int p[6] = {1, 2, 3, 4, 5, 6};
 
    do {
        bool ok = true;
        for (int i = 0; i < 6; i++) {
            int tmp = 0;
            for (int j = 0; j < 6; j++) {
                if(p[j] == a[i][j]){
                    tmp++;
                }
            }
            if (tmp != cnt[i]) {
                ok = false;
                break;
            }
        }
        if (ok) {
            for (int i = 0; i < 6; i++) cout << p[i] << " \n"[i == 5];
            break;
        }
    }
    while (next_permutation(p, p + 6)); // 所有排列可能 (字典序 小~大)
 
}