?題目傳送門：https://www.luogu.org/problemnew/show/P1236

這個題dalao都是用后綴表達式做的，而我這種蒟蒻什么都不會只能強行爆搜失敗。思路簡單：假設輸入的四個數為6 6 6 6，那么我們可以看出答案可以為6+6+6+6=24或6*6-6-6=24，那么無論如何式子都是由三個運算符號和四個數字組成的，那么在dfs（i）中我們根據i的奇偶來判斷枚舉的是什么（奇數位數字，偶數位運算符最后i==8時判斷）。注意點：題目中要求+或*時兩數若有大小先輸大小（頭腦發(fā)昏的我開始-和/都交換了位置），不要忘記了1 3 5 7，這種由兩部分分別運算最后合起來的式子（3-1=2,7+5=12,12*2=24）。

那么最后來解釋一下為什么說是偽題解，以下程序它本身非常神奇地無法處理輸入的四個數中有重復的數且無解的情況（貌似是棧溢出了但我弄了兩個小時也沒弄好），直到我火了，我開始用一個sum1判斷這個輸入有無重復的數，再用sum規(guī)定在有重復的數（即sum1=1）有一個搜索底線（不太好用深度或廣度來形容），簡單的說，就是讓它在棧溢出的邊緣瘋狂試探，搜索時不再死循環(huán)。以下為代碼（這個300多行的代碼令人神魂顛倒盡管很多是復制的）：

#include<stdio.h> #include<stdlib.h> int a[100]={0},step1[100]={0},step2[100]={0},vis[1000]={0}; int flag=0,sum=0,sum1=0; int judge1() {int i,ans;ans=0;for(i=1;i<=3;i++)if(step2[i]==1){if(i==1)ans=step1[i]+step1[i+1];else ans+=step1[i+1];}else if(step2[i]==2){if(i==1)ans=step1[i]-step1[i+1];else ans-=step1[i+1];}else if(step2[i]==3){if(i==1)ans=step1[i]*step1[i+1];else ans*=step1[i+1];}else if(step2[i]==4){if(step1[i]%step1[i+1]!=0) return 0;if(i==1)ans=step1[i]/step1[i+1];else ans/=step1[i+1];}if(ans==24)return 1;return 0; } int judge2() {int i,j,ans1=0,ans2=0;if(step2[1]==1){ans1=step1[1]+step1[2];}else if(step2[1]==2){ans1=step1[1]-step1[2];}else if(step2[1]==3){ans1=step1[1]*step1[2];}else if(step2[1]==4){if(step1[1]%step1[2]!=0) return 0;ans1=step1[1]/step1[2];}if(step2[3]==1){ans2=step1[3]+step1[4];}else if(step2[3]==2){ans2=step1[3]-step1[4];}else if(step2[3]==3){ans2=step1[3]*step1[4];}else if(step2[3]==4){if(step1[3]%step1[4]!=0) return 0;ans2=step1[3]/step1[4];}if(step2[2]==1)ans1=ans1+ans2;else if(step2[2]==2)ans1=ans1-ans2;else if(step2[2]==3)ans1=ans1*ans2;else if(step2[2]==4){if(ans1%ans2!=0)return 0;ans1=ans1/ans2;}if(ans1==24) return 1;return 0; } int write1() {int i,j,ans=0,tmp;if(step2[1]==1){ans=step1[1]+step1[2];if(step1[1]>step1[2])printf("%d+%d=%d\n",step1[1],step1[2],ans);else printf("%d+%d=%d\n",step1[2],step1[1],ans);}else if(step2[1]==2){ans=step1[1]-step1[2];if(step1[1]>step1[2])printf("%d-%d=%d\n",step1[1],step1[2],ans);else printf("%d-%d=%d\n",step1[2],step1[1],ans);}else if(step2[1]==3){ans=step1[1]*step1[2];if(step1[1]>step1[2])printf("%d*%d=%d\n",step1[1],step1[2],ans);else printf("%d*%d=%d\n",step1[2],step1[1],ans);}else if(step2[1]==4){ans=step1[1]/step1[2];if(step1[1]>step1[2])printf("%d/%d=%d\n",step1[1],step1[2],ans);else printf("%d/%d=%d\n",step1[2],step1[1],ans);}for(i=2;i<=3;i++)if(step2[i]==1){tmp=ans;ans+=step1[i+1];if(tmp>step1[i+1])printf("%d+%d=%d\n",tmp,step1[i+1],ans);else printf("%d+%d=%d\n",step1[i+1],tmp,ans);}else if(step2[i]==2){tmp=ans;ans-=step1[i+1];printf("%d-%d=%d\n",tmp,step1[i+1],ans);}else if(step2[i]==3){tmp=ans;ans*=step1[i+1];if(tmp>step1[i+1])printf("%d*%d=%d\n",tmp,step1[i+1],ans);else printf("%d*%d=%d\n",step1[i+1],tmp,ans);}else if(step2[i]==4){tmp=ans;ans/=step1[i+1];printf("%d/%d=%d\n",tmp,step1[i+1],ans);}return 0; } int write2() {int i,ans1=0,ans2=0;if(step2[1]==1){ans1=step1[1]+step1[2];if(step1[1]>step1[2])printf("%d+%d=%d\n",step1[1],step1[2],ans1);else printf("%d+%d=%d\n",step1[2],step1[1],ans1);}else if(step2[1]==2){ans1=step1[1]-step1[2];if(step1[1]>step1[2])printf("%d-%d=%d\n",step1[1],step1[2],ans1);else printf("%d-%d=%d\n",step1[2],step1[1],ans1);}else if(step2[1]==3){ans1=step1[1]*step1[2];if(step1[1]>step1[2])printf("%d*%d=%d\n",step1[1],step1[2],ans1);else printf("%d*%d=%d\n",step1[2],step1[1],ans1);}else if(step2[1]==4){ans1=step1[1]/step1[2];printf("%d/%d=%d\n",step1[1],step1[2],ans1);}if(step2[3]==1){ans2=step1[3]+step1[4];if(step1[3]>step1[4])printf("%d+%d=%d\n",step1[3],step1[4],ans2);else printf("%d+%d=%d\n",step1[4],step1[3],ans2);}else if(step2[3]==2){ans2=step1[3]-step1[4];printf("%d-%d=%d\n",step1[3],step1[4],ans2);}else if(step2[3]==3){ans2=step1[3]*step1[4];if(step1[3]>step1[4])printf("%d*%d=%d\n",step1[3],step1[4],ans2);else printf("%d*%d=%d\n",step1[4],step1[3],ans2);}else if(step2[3]==4){ans2=step1[3]/step1[4];printf("%d/%d=%d\n",step1[3],step1[4],ans2);}if(step2[2]==1){if(ans1>ans2)printf("%d+%d=24\n",ans1,ans2);else printf("%d+%d=24\n",ans2,ans1);}else if(step2[2]==2){if(ans1>ans2)printf("%d-%d=24\n",ans1,ans2);else printf("%d-%d=24\n",ans2,ans1);}else if(step2[2]==3){if(ans1>ans2)printf("%d*%d=24\n",ans1,ans2);else printf("%d*%d=24\n",ans2,ans1);}else if(step2[2]==4){if(ans1>ans2)printf("%d/%d=24\n",ans1,ans2);else printf("%d/%d=24\n",ans2,ans1);}return 0; } int dfs(int i) {int j;sum++;if(i==8){if(judge1()==1){flag=1;write1();}return 0;}if(flag==1)return 0;if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;if(i%2==1){for(j=1;j<=4;j++)if(vis[j]==0){step1[i/2+1]=a[j];vis[j]=1;dfs(i+1);if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;step1[i/2+1]=0;vis[j]=0;}}else{for(j=1;j<=4;j++){step2[i/2]=j;dfs(i+1);if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;step2[i/2]=0;}}return 0; } int dfs2(int i) {int j;sum++;if(i==8){if(judge2()==1){flag=1;write2();}return 0;}if(flag==1)return 0;if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;if(i%2==1){for(j=1;j<=4;j++)if(vis[j]==0){step1[i/2+1]=a[j];vis[j]=1;dfs2(i+1);if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;step1[i/2+1]=0;vis[j]=0;}}else{for(j=1;j<=4;j++){step2[i/2]=j;dfs2(i+1);if(sum>3700&&sum1==1) return 0;if(sum>5000) return 0;step2[i/2]=0;}}return 0; } int main() {int n,i;for(i=1;i<=4;i++){scanf("%d",&a[i]);if(vis[a[i]]==0)vis[a[i]]=1;else sum1=1;}memset(vis,0,sizeof(vis));dfs(1);if(flag==0){sum=0;dfs2(1);if(flag==0)printf("No answer!");}return 0; }

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