Jobdu-1531-货币面值[解题代码] | Acm之家


九度-1531-货币面值[解题代码] | Acm之家
小虎是游戏中的一个国王,在他管理的国家中发行了很多不同面额的纸币,用这些纸币进行任意的组合可以在游戏中购买各种装备来提升自己。有一天,他突然很想知道这些纸币的组合不能表示的最小面额是多少,请聪明的你来帮助小虎来解决这个财政问题吧。
输入:
输入包含多个测试用例,每组测试用例的第一行输入一个整数N(N<=100)表示流通的纸币面额数量,第二行是N个纸币的具体表示面额,取值[1,100]。
输出:
对于每组测试用例,输出一个整数,表示已经发行的所有纸币都不能表示的最小面额(已经发行的每个纸币面额最多只能使用一次,但面值可能有重复)。
样例输入:
5 1 2 3 9 100 5 1 2 4 9 100 5 1 2 4 7 100
样例输出:
7 8 15
http://www.cnblogs.com/liushaobo/p/4373759.html
  1. int ZeroOnePack(){  
  2.     int i, j;  
  3.     memset(dp, 0sizeof(dp));  
  4.     for (i = 1; i <= N; ++i){  
  5.         for (j = max; j >= value[i]; --j){  
  6.             dp[j] = Max(dp[j], dp[j-value[i]] + value[i]);  
  7.         }  
  8.     }  
  9.     for (i = 1; i <= max; ++i)  
  10.         if (dp[i] != i)  
  11.             return i;  
  12. }
  1. //动态规划的思想, 对于从第1个到第i个数的和total,  
  2. //如果第i+1个数大于total+1则不会组成total+1.
  3.   
  4.     while (scanf("%d", &N) != EOF){  
  5.         for (i = 0; i < N; ++i){  
  6.             scanf("%d", &value[i]);  
  7.         }  
  8.         qsort(value, N, sizeof(int), Compare);  
  9.         ans = 0;  
  10.         for (i = 0; i < N; ++i){  
  11.             if (value[i] > ans + 1){  
  12.                 break;  
  13.             }  
  14.             else  
  15.                 ans += value[i];  
  16.         }  
  17.         printf("%d\n", ans + 1);  
  18.     } 
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