poj3187(next_permutation实现全排列+杨辉三角)

Backward Digit Sums

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 5752 Accepted: 3335

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:

3   1   2   4

4   3   6

7   9

16

Behind 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.

Write a program to help FJ play the game and keep up with the cows.

Input

Line 1: Two space-separated integers: N and the final sum.

Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically(辞典编纂的) least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4





题解：这是一个倒的杨辉三角，第i个数前面的系数为C(n-1,i-1)，所以sum=a1*C(n-1,0)+a2*C(n-1,1)+……+ai*C(n-1,i-1)+……然后运用next_permutation进行全排列，加上限制条件就OK了

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int comb(int n,int m){//求组合数C(n,m) ,手写的
if(m==0||n==m)return 1;
if(m==1)return n;
int i,sum1=1,sum2=1;
for(i=1;i<=m;i++)sum1*=i;
for(i=n-m+1;i<=n;i++)sum2*=i;
return sum2/sum1;
}
int main(){
int n,sum,i,cnt;
int a[20],s[20];
while(scanf("%d %d",&n,&sum)!=EOF){
for(i=0;i<n;i++){
if(i>(n-1)/2){
s[i]=s[n-1-i];
}
else{
s[i]=comb(n-1,i);
}
}

for(i=0;i<n;i++)a[i]=i+1;
do{
cnt=0;
for(i=0;i<n;i++){
cnt+=a[i]*s[i];
}
if(cnt==sum){
for(i=0;i<n-1;i++)printf("%d ",a[i]);
printf("%d\n",a[n-1]);
break;
}
}while(next_permutation(a,a+n));

}
return 0;
}