摘要：#include <stdio.h>#define MAX 100int len, sum;char list[MAX];static int perm(int k, int m);static int swap(int i, int j);int main(){ scanf("%s",list); len = strlen(list); sum = perm(0, len-1);//not include '/n' printf("sum = %d/n", sum); return 0;}static int perm(in
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摘要：Backward Digit SumsTime Limit:1000MSMemory Limit:65536KDescriptionFJ 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 sing
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摘要：第一次用G++提交TLE，第二次用C++ 460ms水过！标记排列Time Limit:1000MSMemory Limit:30000KTotal Submissions:9419Accepted:4168Description题目描述：大家知道，给出正整数n，则1到n这n个数可以构成n！种排列，把这些排列按照从小到大的顺序（字典顺序）列出，如n=3时，列出1 2 3，1 3 2，2 1 3，2 3 1，3 1 2，3 2 1六个排列。任务描述：给出某个排列，求出这个排列的下k个排列，如果遇到最后一个排列，则下1排列为第1个排列，即排列1 2 3…n。比如：n = 3，k=2 给出排列2 3
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摘要：本来想用c实现，可是还没来得及细细研究，这里用STL中的next_permutation水过标记一下，以后自己C实现题目地址http://poj.org/problem?id=1731OrdersTime Limit:1000MSMemory Limit:10000KDescriptionThe stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter are st
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