提交记录 #362

编译情况：编译成功

code.c: In function ‘main’:
code.c:109:16: warning: ‘sig’ may be used uninitialized in this function [-Wmaybe-uninitialized]
  109 |         else if(sig=='-')
      |                ^

固定测试点#1:

Wrong Answer

0.000s

232 kB

固定测试点#2:

Accepted

0.000s

236 kB

固定测试点#3:

Accepted

0.000s

232 kB

附加测试点暂不可用

不可用

N/A

H20【选做▪大学】高精度加减法

#include<stdio.h>
#include<string.h>
char num1[502],num2[502];int ans[100][502]={0},J,K;
int plus(int j)
{
    int i;
    for(i=501;num1[i]!='\0'||num2[i]!='\0';i--)
    {
        ans[j][i]=ans[j][i]+num1[i]-48+num2[i]-48;
        if(ans[j][i]>=10)
        {
            ans[j][i-1]=ans[j][i-1]+1;ans[j][i]=ans[j][i]-10;
        }
    }
    return i+1;
}
int minus(int j)
{
    int i;
    for(i=501;num1[i]!=48||num2[i]!=48;i--)
    {
        ans[j][i]=ans[j][i]+num1[i]-num2[i];
        if(ans[j][i]<0)
        {
            ans[j][i-1]=ans[j][i-1]-1;ans[j][i]=ans[j][i]+10;
        }
    }
    return i+1;
}
int minusanti(int j)
{
    int i;
    for(i=501;num1[i]!=48||num2[i]!=48;i--)
    {
        ans[j][i]=ans[j][i]+num2[i]-num1[i];
        if(ans[j][i]<0)
        {
            ans[j][i-1]=ans[j][i-1]-1;ans[j][i]=ans[j][i]+10;
        }
    }
    return i-1;
}
int max(int x,int y)
{
    if(x>y)
    {
        return x;
    }
    else
    {
        return y;
    }
}
int main()
{
    char sig;int n,i,j,k,d;
    for(j=0;j<502;j++)
        {
            num1[j]='0';
        }
        for(j=0;j<502;j++)
        {
            num2[j]='0';
        }
    scanf("%d\n",&n);
    for(i=0;i<n;i++)
    {
        for(j=0;j<502;j++)
        {
            scanf("%c",&num1[j]);
            if(num1[j]=='+'||num1[j]=='-')
            {
                sig=num1[j];num1[j]='0';j=j-1;break;
            }
        }
        for(k=0;k<502;k++)
        {
            scanf("%c",&num2[k]);
            if(num2[k]=='\n')
            {
                num2[k]='0';k=k-1;break;
            }
        }
        J=j;K=k;//printf("J=%d,K=%d\n",J,K);printf("max=%d\n",max(J,K));
        d=501-J;
        for(j=501;j>=d;j--)
        {
            num1[j]=num1[j-d];
        }
        for(;j>=0;j--)
        {
            num1[j]='0';
        }
        d=501-K;
        for(k=501;k>=d;k--)
        {
            num2[k]=num2[k-d];
        }
        for(;k>=0;k--)
        {
            num2[k]='0';
        }
        if(sig=='+')
        {
            //printf("%c\n",sig);
            plus(i);
        }
        else if(sig=='-')
        {
            if(J>K)
            {
                minus(i);
            }
            else if(J<K)
            {
                minusanti(i);ans[i][0]=-1;
            }
            else
            {
                for(k=501-max(J,K);k<502;k++)
                {
                    if(num1[k]>num2[k])
                    {
                        minus(i);
                    }
                    else if(num1[k]<num2[k])
                    {
                        minusanti(i);ans[i][0]=-1;
                    }
                }
            }
        }
    }
    for(i=0;i<n;i++)
    {
        if(ans[i][0]==-1)
        {
            printf("-");
        }
        for(j=1;j<502;j++)
        {
            if(ans[i][j]!=0||j==501)
            {
                for(;j<502;j++)
                {
                    printf("%d",ans[i][j]);
                }
            }
        }
        printf("\n");
    }
    return 0;
}