hdoj 1709 The Balance(母函数)
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因为是天平,所以组合之间的差值也都是可以得到的,这差值也是新的组合数。
代码:
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1e4+5;int n, c1[maxn], c2[maxn], num[maxn], ans[maxn];int main(void){ while(cin >> n) { int sum = 0; for(int i = 1; i <= n; i++) scanf("%d", &num[i]), sum += num[i]; memset(c1, 0, sizeof(c1)); memset(c2, 0, sizeof(c2)); c1[0] = c1[num[1]] = 1; for(int i = 2; i <= n; i++) { for(int j = 0; j <= sum; j++) for(int k = 0; k+j <= sum && k <= num[i]; k += num[i]) { if(k > j) c2[k-j] += c1[j]; else c2[j-k] += c1[j]; c2[k+j] += c1[j]; } for(int j = 0; j <= sum; j++) c1[j] = c2[j], c2[j] = 0; } int count = 0; for(int i = 1; i <= sum; i++) if(!c1[i]) ans[count++] = i; if(!count) puts("0"); else { printf("%d\n", count); for(int i = 0; i < count; i++) printf("%d%c", ans[i], i==count-1? '\n' : ' '); } } return 0;}
The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7615 Accepted Submission(s): 3165
Problem Description
Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.
Input
The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.
Output
For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.
Sample Input
31 2 439 2 1
Sample Output
024 5
Source
HDU 2007-Spring Programming Contest
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