1 条题解

  • 0
    @ 2021-6-15 12:55:34

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    long long a[101010],cnt=1,sum=0;
    char c[101010];
    int main()
    {
        cin>>a[cnt];
        a[cnt]%=10000;
        while(cin>>c[cnt])
        {
                cin>>a[++cnt];
            a[cnt]%=10000;
        }
        for(int i=cnt;i>=1;i--)
        {
            if(c[i]=='*')
            {
                a[i]*=a[i+1];
                a[i]%=10000;
                a[i+1]=0;
            }
        }
        for(int i=1;i<=cnt;i++)
        {
            sum+=a[i];
            sum%=10000;
        }     
        cout<<sum;
    }
    

    Pascal :

    var a:array [1..100001] of integer;
        b:array [1..100001] of char;
        s:string;z:char;h,i,j:longint;
    begin
      while not eof do
        begin
          read(z);
          s:='';
          while z in ['0'..'9'] do
            begin
              s:=s+z;
              read(z);
            end;
          i:=i+1;
          val(s,j);
          a[i]:=j mod 10000;
          b[i]:=z;
        end;
      for j:=1 to i do
      if b[j]='*' then
        begin
          a[j+1]:=a[j]*a[j+1] mod 10000;
          a[j]:=0;
        end;
      for j:=1 to i do h:=h+a[j];
      write(h mod 10000);
    end.
    
    • 1

    信息

    ID
    700
    时间
    1000ms
    内存
    128MiB
    难度
    10
    标签
    递交数
    4
    已通过
    1
    上传者