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3 1 2 4Behind FJ‘s back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ‘s mental arithmetic capabilities.
4 3 6
7 9
16
输入
输出
样例输入
4 16
样例输出
3 1 2 4
提示
题意：
我们把1-n这n个数的某个排列摆成一排，然后相邻两个数的和放到下一行，依此类推，形成一个三角形
3 1 2 4
4 3 6
7 9
16
给你n和最终得到的和k，求第一行的系列（字典序最小）
我们可以发现每个数字的计算次数是一个杨辉三角
1 1
1 2 1
1 3 3 1
1 4 6 4 1
......
所以我们只要枚举1-n的全排列，计算sum{now[i]*as(n-1,i)}是否为K就行了
#include <iostream> #include <queue> #include <cmath> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; typedef long long LL; const LL INF = 0xfffffff; const int maxn = 500; const LL MOD = 1e9+7; int ok[maxn], now[maxn], as[maxn][maxn], n, m, vis[maxn], flag;
///poj 3187 void init() { int i, j; as[0][0] = 1; for(i = 0; i <= 10; i++) { as[i][0] = as[i][i] = 1; for(j = 1; j < i; j++) as[i][j] = as[i-1][j] + as[i-1][j-1]; } } void dfs(int cnt) { if(flag )return ; if(cnt == n) { int ans = 0; for(int i = 0; i < n; i++) ans += as[n-1][i] * now[i]; if(ans == m) { flag = 1; for(int i = 0; i < n; i++) ok[i] = now[i]; } return ; } for(int i = 1; i <= n; i++) { if(!vis[i]) { vis[i] = 1; now[cnt++] = i; dfs(cnt); vis[i] = 0; cnt--; } } // return ; } int main() { init(); while(~scanf("%d %d", &n, &m)) { flag = 0; memset(vis, 0, sizeof(vis)); dfs(0); printf("%d", ok[0]); for(int i = 1; i < n; i++) printf(" %d", ok[i]); printf("\n"); } return 0; }
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原文地址：http://www.cnblogs.com/PersistFaith/p/4832911.html
