蓝桥杯加减乘

秒速五厘米 2022-05-29 07:20 113阅读 0赞

请对于下面式子进行填空，填入加减乘，使这个表达式成立。

11 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 22 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 33 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 44 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 55 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 66 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 77 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 88 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 99 [![ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png][]][ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png] 1010 == 00

请输出一共有多少种方案可以使得表达式成立。

思路：把所有表达式情况枚举出来，然后计算其结果是否为0

#include<cstdio>
    #include<iostream>
    #include<string>
    #include<cstdlib>
    #include<algorithm>
    using namespace std;
    int ans,sum;
    char ss[11]={'0','1','2','3','4','5','6','7','8','9','1'};
    void dfs(string str,int s,int cnt,int x,int y,int z)
    {     
         if(cnt>=11)
         {
            if(x>=1&&y>=1&&z>=1)//x,y,z分别表示加号，减号，乘号的个数，题目的意思是三种符号必须都要有 
            {
                int s0=1,ok=1,flag=1;
                int a[20],l=0;
                str=str+'+';//str字符串就代表所有的表达式情况，然后再从里面找等式成立的个数 
                //cout<<str<<endl;
                for(int i=1;i<20;i+=2)//计算表达式结果 
                {
                        if(str[i]!='*')
                        {
                              if(str[i-2]=='-'){
                              s0=-1;
                              }
                              s0=ok*s0*(str[i-1]-48);
                              if(i==19)
                              s0=s0*10;
                              a[l++]=s0;
                              s0=1;
                              ok=1;
                        }                                         
                        else
                        {
                             if(str[i-2]=='-'&&ok==1)
                                 ok=-1;
                             s0=s0*(str[i-1]-48);
                        }
                }
                s0=0;
                //用a[i]数组把等式的中间结果存起来，方便运算 
                //比如 1-2*3*4+5-6*7*8+9-10；
                //用a[0]=1,a[1]=-120,a[2]= -6*7*8,a[3]=9,a[4]=-10
                for(int i=0;i<l;i++)
                {
                     //printf("%d ",a[i]);
                     s0=s0+a[i];       
                }
               // printf("%d\n",s0);
                if(s0==0) // 结果为0就加1  
                sum++;
             }
             return ;
         }
         //像个三叉树 
         dfs(str+'+'+ss[cnt],s+cnt,cnt+1,x+1,y,z);//如果有 加号  ，x加1，下同 
         dfs(str+'-'+ss[cnt],s-cnt,cnt+1,x,y+1,z); 
         dfs(str+'*'+ss[cnt],s*cnt,cnt+1,x,y,z+1);     
    }
    int main(){
        ans=0;
        sum=0;
        string s="1";//反正1总是在最前面的 
        dfs(s,1,2,0,0,0);
        printf("%d\n",sum);
        system("PAUSE");
        return EXIT_SUCCESS;
    }

#include<cstdio>
    #include<iostream>
    #include<string>
    #include<cstdlib>
    #include<algorithm>
    using namespace std;
    char ss[11]={'0','1','2','3','4','5','6','7','8','9','1'};
    int dfs(string str,int s,int cnt,int x,int y,int z){
         
         
         if(cnt>=11)
         {
            if(x>=1&&y>=1&&z>=1)
            {
                int s0=1,ok=1;
                int a[20],l=0;
                str=str+'+';
                //cout<<str<<endl;
                for(int i=1;i<20;i+=2)
                {
                        if(str[i]!='*')
                        {
                              if(str[i-2]=='-'){
                              s0=-1;
                              }
                              s0=ok*s0*(str[i-1]-48);
                              if(i==19)
                              s0=s0*10;
                              a[l++]=s0;
                              s0=1;
                              ok=1;
                        }                                         
                        else
                        {
                             if(str[i-2]=='-'&&ok==1)
                                 ok=-1;
                             s0=s0*(str[i-1]-48);
                        }
                }
                s0=0;
                for(int i=0;i<l;i++)
                {
                        //printf("%d ",a[i]);
                     s0=s0+a[i];       
                }
               // printf("%d\n",s0);
                if(s0==0)   
                return 1;
                else
                return 0;
             }
             return 0;
         }
         int sum=0;
         sum+=dfs(str+'+'+ss[cnt],s+cnt,cnt+1,x+1,y,z);
         sum+=dfs(str+'-'+ss[cnt],s-cnt,cnt+1,x,y+1,z);
         sum+=dfs(str+'*'+ss[cnt],s*cnt,cnt+1,x,y,z+1);
         return sum;      
    }
    int main(){
        string s="1";
        int t=dfs(s,1,2,0,0,0);
        printf("%d\n",t);
        system("PAUSE");
        return EXIT_SUCCESS;
    }

[ba153f0b44c094ff3b7db7ff16f88b3230dfe815.png]: /images/20220529/fdbbe606aeaa4ec2ae9c3282531001d3.png