sdut1642Simple Arithmetics(模拟)

链接

发个长长的模拟 这题要注意的地方挺多 -的个数 以及对齐的情况 全都注意好了 大数的加减乘就可以了

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 using namespace std;
 #define N 510
 char s1[N],s2[N];
 int s[N][N*],o[N];
 int main()
 {
     int t,i,j,k1,k2,len;
     char c,tc;
     scanf("%d",&t);
     getchar();
     while(t--)
     {
         len = ;k1=;
         for(;;)
         {
            scanf("%c",&c);
            if(c=='+'||c=='-'||c=='*')
            {
                tc = c;
                s1[k1] = '\0';
                break;
            }
            s1[k1++] = c;
         }
         gets(s2);
         k2 = strlen(s2);
         for(i =  ; i <= k2; i++)
             for(j =  ; j <= k2+k1+ ; j++)
             s[i][j] = ;
         for(i =  ; i <= k2 ; i++)
         o[i] = ;
         if(tc=='+')
         {
             i = k1-,j=k2-;
             int g = ;
             while(i>=&&j>=)
             {
                 int ko = s[][g]+s1[i--]-''+s2[j--]-'';
                 s[][g+] = ko/;
                 s[][g++] = ko%;
             }
             while(i>=)
             {
                 int ko = s[][g]+s1[i--]-'';
                 s[][g+] = ko/;
                 s[][g++] = ko%;
             }
             while(j>=)
             {
                 int ko = s[][g]+s2[j--]-'';
                 s[][g+] = ko/;
                 s[][g++] = ko%;
             }
             g++;
             while(s[][g]==&&g>) g--;
             int dd = max(g+,max(k1,k2+));
             for(i = ; i <= dd-k1 ; i++) printf(" ");
             printf("%s\n",s1);
             for(i = ; i <= dd--k2 ; i++) printf(" ");
             printf("%c%s\n",tc,s2);
             for(i = ; i <= dd-max(k2+,g+) ; i++) printf(" ");
             for(i = ; i <= max(k2+,g+) ; i++)
             printf("-");
             puts("");
             for(i = ; i <= dd--g ; i++) printf(" ");
             for(i = g ; i >=  ; i--)
             printf("%d",s[][i]);
             puts("");
         }
         else if(tc=='-')
         {
             i = k1-,j=k2-;
             int g = ;
             while(i>=&&j>=)
             {
                 int ko = s[][g]+s1[i]-s2[j];
                 if(ko<)
                 {
                     ko+=;s[][g+]--;
                 }
                 s[][g++] = ko;i--;j--;
             }
             while(i>=)
             {
                 int ko = s[][g]+s1[i]-'';
                 if(ko<)
                 {
                     ko+=;s[][g+]--;
                 }
                 s[][g++] = ko;i--;
             }
             while(s[][g]==&&g>) g--;
             int dd = max(k1,k2+);
             for(i = ; i <= dd-k1 ; i++) printf(" ");
             printf("%s\n",s1);
             for(i = ; i <= dd--k2 ; i++) printf(" ");
             printf("%c%s\n",tc,s2);
             for(i = ; i <= dd-max(k2+,g+) ; i++) printf(" ");
             for(i = ; i <= max(k2+,g+) ; i++) printf("-");
             puts("");
             for(i = ; i <= dd--g ; i++) printf(" ");
             for(i = g ; i >=  ; i--)
             printf("%d",s[][i]);
             puts("");
         }
         else
         {
             for(i = k2- ; i >=  ; i--)
             {
                 o[i] = k2--i;
                 for(j = k1- ; j >=  ; j--)
                 {
                     int ko = s[i][o[i]]+(s2[i]-'')*(s1[j]-'');
                     s[i][o[i]+] = ko/;
                     s[i][o[i]++] = ko%;
                 }
                 while(s[i][o[i]]==&&o[i]>(k2--i)) o[i]--;
                 for(j =  ; j <= max(o[k2],o[i]) ; j++)
                 {
                     int ko = s[k2][j]+s[i][j];
                     s[k2][j] = ko%;
                     s[k2][j+] += ko/;
                 }
                 o[k2] = max(o[k2],o[i]);
                 if(s[k2][o[k2]+]) o[k2]++;
             }
             while(s[k2][o[k2]]==&&o[k2]>) o[k2]--;
             int dd = max(max(k1,k2+),o[k2]+);
             int td = max(k2+,o[k2-]+);
             for(i = ; i <= dd-k1 ; i++) printf(" ");
             printf("%s\n",s1);
             for(i = ; i <= dd-k2- ; i++) printf(" ");
             printf("%c%s\n",tc,s2);
             for(i = ; i <= dd-td ; i++) printf(" ");
             for(i = ; i <= td ; i++)
             printf("-");puts("");
             for(i = k2- ; i >=  ; i--)
             {
                 for(j = ; j < dd-o[i] ; j++) printf(" ");
                 for(j = o[i] ; j >= k2-i- ; j--) printf("%d",s[i][j]);
                 puts("");
             }
             if(k2>)
             {
                 for(i = ; i < dd-o[k2] ; i++) printf(" ");
                 for(i = ; i <= o[k2]+ ; i++)
                 printf("-");puts("");
                 for(i = ; i < dd-o[k2] ; i++) printf(" ");
                 for(i = o[k2] ; i >=  ; i--)
                 printf("%d",s[k2][i]);
                 puts("");
             }
         }
         puts("");
     }
     return ;
 }

 /**************************************
     Problem id    : SDUT OJ 1642
     User name    : shang
     Result        : Accepted
     Take Memory    : 912K
     Take Time    : 160MS
     Submit Time    : 2014-02-15 20:53:30
 **************************************/