NYOJ-组合数

NYOJ-组合数

本来思路是dfs，不过可以利用C++的STL可以解决
C++ STL

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int n, r;
bool isD(string s) {
    char c1 = s[0];
    for (int i = 1; i < r; i++) {
        if (c1 < s[i]) return false;
        c1 = s[i];
    }
    return true;
}
int main() {
    //freopen("in.txt", "r", stdin);
    string s1;
    scanf("%d%d", &n, &r);
    for (int i = n; i > 0; i--) {
        s1 += '0' + i;
    }
    string s2;
    while(prev_permutation(s1.begin(), s1.end())) {
        if (isD(s1) && s2 != s1.substr(0, r)){
            s2 = s1.substr(0, r);
            cout << s2 << endl;
        }    
    }

    return 0;
}

dfs

#include<cstdio>
#include<cstdlib>
int box[11], book[11], n;
void dfs(int step, int count) {

    if (count >= n+1) {
        for (int i = 0; i < n; i++) {
            printf("%d",box[i]);
        }
        puts("");
        if (box[0] == n) exit(-1);
        return;
    }

    // 尝试处理每一种可能
    for (int i = step; i > 0; i--) {
        box[count-1] = step--;
        dfs(step, count+1);
    } 


}

int main() {
//  freopen("in.txt","r",stdin);     //从in.txt 中读入数据 
    int m;
    scanf("%d%d", &m, &n);
    dfs(m, 1);
    return 0;
}        

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