POJ 3187 Backward Digit Sums

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1.题目描述：点击打开链接

2.解题思路：本题通过观察发现实际上是找符合这样的等式的一个排列：

C(n-1,0)*a[0]+C(n-1,1)*a[1]+...+C(n-1,n-1)*a[n-1]==sum

其中数组a是1~n的一个排列，C(n-1,i)代表组合数。这样的话，可以花费O(N^2)时间预先计算好所有组合数，然后用next_permutation函数枚举下一个排列即可。如果发现正好等于sum，停止枚举。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 11int a[N];int c[N][N];int n, sum;void init(){for (int i = 1; i < N; i++)c[i][0] = c[i][i] = 1;for (int i = 2; i < N;i++)for (int j = 1; j < i; j++)c[i][j] = c[i - 1][j-1] + c[i - 1][j];}int main(){//freopen("t.txt", "r", stdin);init();while (~scanf("%d%d", &n, &sum)){for (int i = 0; i < n; i++)a[i] = i + 1;do{int res = 0;for (int i = 0; i < n; i++)res += c[n - 1][i] * a[i];if (res == sum)break;} while (next_permutation(a , a + n));for (int i = 0; i < n; i++)printf("%d%c", a[i], i == n - 1 ? '\n' : ' ');}return 0;}

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