Sicily 1050 Numbers & Letters(DFS)

//DFS回溯，其实就是24点游戏的问题//没想到那么蛋疼，居然卡了我很久，显然的深搜题//回溯思路是这样的，5个数，先任意找2个数进行加减乘除，把这2个数运算后的结果当做一个数，按相同的方法搜下去//把4个数任取2个然后合并成3个，再继续搜下去//搜索树的深度为5， 第一层最多有C(2,5)*5种分支,这里5表示两个数做运算最多能产生5种结果//第二层C(2,4)*5...显然搜索树规模挺小的，加上除法这一运算产生的几率更小，因此这样搜过去必然没问题//我WA了很久的一个原因是初始化，没有把ANS初始化为-INF是最大的失败..#include<iostream>#include<cstdio>#include<algorithm>using namespace std;bool ok;int T;long A[5],target,ans;long add(long x,long y){return x + y;}long sub(long x,long y){return x - y;}long mul(long x,long y){return x * y;}long Div(long x,long y){if(x < y)swap(x,y);if(y == 0 || x % y != 0)return -1;return x / y;}void dfs(long N[],int size){if(ok)return;if(N[0] <= target && N[0] > ans){ans = N[0];if(ans == target){ok = 1;return;}}if(size == 1)return;long num[5];for(int i = 0;i < size;++i){for(int j = i+1;j < size;++j){for(int k = 0,h = 1;k < size;++k)if(k != i && k != j)num[h++] = N[k];//填补除去选中的I,J两个数num[0] = add(N[i],N[j]);dfs(num,size-1);num[0] = sub(N[i],N[j]);dfs(num,size-1);num[0] = -num[0];dfs(num,size-1);//减法有可以交换num[0] = mul(N[i],N[j]);dfs(num,size-1);num[0] = Div(N[i],N[j]);if(num[0] != -1)dfs(num,size-1);}}}int main(){//freopen("in.txt","r",stdin);scanf("%d",&T);while(T--){ok = 0;for(int i = 0;i < 5;++i)scanf("%ld",&A[i]);scanf("%ld",&target);ans = -2000000000;for(int i = 0;i < 5;++i)if(A[i] <= target && A[i] > ans)ans = A[i];dfs(A,5);printf("%ld/n",ans);}}