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2016蓝桥杯省赛C/C++A组第六题 寒假作业

题意:现在小学的数学题目也不是那么好玩的。 
看看这个寒假作业: 
□ + □ = □ 
□ - □ = □ 
□ × □ = □ 
□ ÷ □ = □ 
每个方块代表1~13中的某一个数字,但不能重复。 
比如: 
6 + 7 = 13 
9 - 8 = 1 
3 * 4 = 12 
10 / 2 = 5 
以及: 
7 + 6 = 13 
9 - 8 = 1 
3 * 4 = 12 
10 / 2 = 5 
就算两种解法。(加法,乘法交换律后算不同的方案) 
你一共找到了多少种方案? 

分析:回溯即可,但是如果等到cur==12再统一判断,时间复杂度会达到2的13次方,非常慢,所以当一个式子填完后立即判断,不满足则立即返回。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 100 + 10;const int MAXT = 10000 + 10;using namespace std;int vis[15];int a[15];int ans;void dfs(int cur){    if(cur == 3){        if(a[0] + a[1] != a[2]) return;    }    if(cur == 6){        if(a[3] - a[4] != a[5]) return;    }    if(cur == 9){        if(a[6] * a[7] != a[8]) return;    }    if(cur == 12){        if(a[10] * a[11] == a[9]){            ++ans;        }        return;    }    for(int i = 1; i <= 13; ++i){        if(vis[i]) continue;        vis[i] = 1;        a[cur] = i;        dfs(cur + 1);        a[cur] = 0;        vis[i] = 0;    }}int main(){    dfs(0);    printf("%d\n", ans);    return 0;}

  

2016蓝桥杯省赛C/C++A组第六题 寒假作业