Shredding Company

http://162.105.81.212/JudgeOnline/problem?id=1416

一开始以为是动态规划，后来以为是贪心，发现都不行，一看数据只有六位，枚举切割方案就行了(至多32种)，代码写得很不顺(模拟功底太弱了)，WA了一次。

#include <iostream> using namespace std ; int ans , m , n , len , b ; char str[10] ; bool rejected , cut[6] ; void Try (int k)//模拟切割 { int i , j , sum , tmp ; cut[len - 1] = true ; for (j = i = sum = 0 ; i < len ; i ++) if (cut[i]) { for (tmp = 0 ; j <= i ; j ++) { tmp *= 10 ; tmp += str[j] - '0' ; } sum += tmp ; if (sum > n) return ; } if (sum > ans) { ans = sum ; b = k ; rejected = false ; } else if (sum == ans) rejected = true ; } void Comp () { int i , j , k ; for (i = 1 ; i < (1 << (len - 1)) ; i ++)//枚举切割方案（用二进制表示） { memset (cut , false , sizeof (cut)) ; for (j = 0 , k = 1 ; k <= i ; k <<= 1 , j ++) if (i & k) cut[j] = true ; Try (i) ; } } int main () { int i , j , sum ; while (scanf ("%d%s" , &n , str) , n , atoi (str)) { m = atoi (str) ; len = (int)strlen (str) ; if (n >= m)//和的最大值(切的块数越多，和越小) { printf ("%d %d/n" , m , m) ;//这里把第一个m搞错成n, WA了一次 continue ; } for (ans = i = sum = 0 ; i < len ; i ++) sum += str[i] - '0' ; if (sum > n)//和的最小值 { printf ("error/n") ; continue ; } Comp () ; if (rejected) { printf ("rejected/n") ; continue ; } printf ("%d " , ans) ; putchar (str[0]) ; for (j = i = 1 ; i <= b ; i <<= 1) { if (b & i) putchar (' ') ; putchar (str[j ++]) ; } while (j < len) putchar (str[j ++]) ; putchar ('/n') ; } return 0 ; }