题目大一:根据题目的算法,找出最小的C。
解题思路:因为题目有说C至少要是w集合的某一个的倍数,又要找最小的,所以现在w内找个最小的,然后在根据所给的条件遍历判度所有的w是否满足要求,如果不满足就根据要求更换C,然后再次判断。
#include<stdio.h> #include<string.h> const int N = 100000; char s[N]; int word[15], n, flag; int min( int a, int b) { return a < b ? a:b; } void handle() { n = 0; memset(word, 0, sizeof(word)); for ( int i = 0; s[i] ; i++) { if (s[i] != ‘ ‘) word[n] = word[n] * 32 + s[i] - ‘a‘ + 1; else if ( s[i] == ‘ ‘ && s[i - 1] != ‘ ‘) n++; } n++; } int cal(int cur) { int i, j; for (i = 1; i <= n; i++) for (j = i + 1; j <= n ;j++) { if ( word[j - 1] == 0 || word[ i - 1] == 0 ) continue; if (cur / word[i - 1] % n == cur / word[ j - 1] % n) return min((cur / word[i - 1] + 1 ) * word[i - 1] , ( cur / word [ j - 1] + 1) * word[j- 1] ); } flag = 1; return cur; } int main() { while (gets(s)) { flag = 0; handle(); puts(s); int max = 999999999; for (int i = 0; i < n; i++ ) if ( word[i] < max) max = word[i]; while (!flag) { max = cal(max); } printf("%d\n\n", max); } return 0; }
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原文:http://blog.csdn.net/u012997373/article/details/22758957