poj 3781

题意：给你n个数。求出n个数的规律和（由相邻的数相加，每次减小一个数，直到只剩1个数）。

这是杨辉三角的性质。

好，接下来的全排列，可以使用STLd的next_permutation。这只比dfs慢一丢丢。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn = 15;int n, ans;int yang[maxn][maxn], num[maxn];void init(){    for(int i = 0; i < maxn; i++)        for(int j = 0; j <= i; j++)        {            if(j == 0 || j == i) yang[i][j] = 1;            else yang[i][j] = yang[i - 1][j] + yang[i - 1][j - 1];        }}bool vis[maxn];bool dfs(int deep, int sum){    //printf("%d %d\n", sum, deep);    if(sum > ans) return 0;    if(deep == n)    {        //printf("%d %d\n", sum, deep);        if(ans == sum) return true;        else return false;    }    for(int i = 1; i <= n; i++)    {        if(!vis[i])        {            vis[i] = 1;            num[deep] = i;            if(dfs(deep + 1, sum + i * yang[n - 1][deep]))            {                //printf("yes\n");                return true;            }            //printf("%d\n", sum);            vis[i] = 0;        }    }    return false;//这个一定要注意}int main(){    init();    while(scanf("%d%d", &n, &ans) == 2)    {        memset(vis, 0, sizeof(vis));        dfs(0, 0);        for(int i = 0; i < n - 1; i++)            printf("%d ", num[i]);        printf("%d\n", num[n - 1]);    }    return 0;}

STL：next_permutation()

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 15;int n, ans;int yang[maxn][maxn], num[maxn];void init(){    for(int i = 0; i < maxn; i++)        for(int j = 0; j <= i; j++)        {            if(j == 0 || j == i) yang[i][j] = 1;            else yang[i][j] = yang[i - 1][j] + yang[i - 1][j - 1];        }}int main(){    init();    while(scanf("%d%d", &n, &ans) == 2)    {        int sum;        for(int i = 0; i <= 10; i++) num[i] = i + 1;        do        {            sum = 0;            for(int i = 0; i < n; i++)            {                sum += num[i] * yang[n - 1][i];                if(sum > ans) break;            }            if(sum == ans) break;        }        while(next_permutation(num, num + n));        //printf("%d\n", sum);        for(int i = 0; i < n - 1; i++)            printf("%d ", num[i]);        printf("%d\n", num[n - 1]);    }    return 0;}