/*
1002 A + B Problem II
Time Limit : 1000 ms Memory Limit : 32768 K Output Limit : 1024 K
VC
*/
//难度在于解决大数运算的方法和实际操作中
#include
#include
#define DIV 4
#define NUMLONG 1000
#define TMAX 20
short deadd[TMAX]={0};
short flag=0;
int ChangetoNum(char * pstr,short * pnum,short now)
{
short ca=0,cb=3;
char divstr[DIV+1];
divstr[DIV]='\0';
if(*pstr=='-')
{
deadd[now]++;
pstr++;
}
while(pstr[ca]!='\0') ca++;
pnum[0]=ca-ca%DIV;
if(pnum[0]==0)//判断是否不足一个保存单元
pnum[0]=1;
else if(pnum[0]==ca)//判断是否没有零碎的保存单元
pnum[0]=(ca-ca%DIV)/DIV;
else
pnum[0]=(ca-ca%DIV)/DIV+1;
//pnum[0]保存数字的保存长度
pnum++;;
while((ca--)>=0)
{
if(cb {
*pnum=(short)atoi(divstr);//分段字符串转换为数字
pnum++;
cb=3;
divstr[cb--]=pstr[ca];
}
else if(ca {
while(cb>=0)
divstr[cb--]='0';
*pnum=(short)atoi(divstr);
}
else
divstr[cb--]=pstr[ca];
}
return 0;
}
int Calculate(short *pnum1,short *pnum2,short *pans)
{
short ca,n,temp;
if(pnum1[0] n=pnum2[0];
else
n=pnum1[0];
for(ca=1;ca {
temp=pnum1[ca]+pnum2[ca]+pans[ca];
if(temp>=10000)//结果数字有进位
{
pans[ca]=temp-10000;
pans[ca+1]=1;
pans[0]=ca+1;
}
else
pans[ca]=temp;
}
pans[0]=pans[0] return 0;
}
int Calculate2(short *pnum1,short *pnum2,short *pans)//pnum1为大数
{
short ca,n,temp;
n=pnum1[0];
pans[0]=n;
for(ca=1;ca {
temp=pnum1[ca]-pnum2[ca]-pans[ca];
if(temp==0 && ca==n)
pans[0]=n-1;
if(temp {
if(pnum1[ca+1]>0)
{
pans[ca]=temp+10000;
pans[ca+1]=1;
}
}
else
pans[ca]=temp;
}
return 0;
}
int PrintAns(char *pleft,char *pright,short *pans,short ca)
{
short cb,temp;
printf("Case %d:\n",ca);
printf("%s + %s = ",pleft,pright);
if(flag==1) //首单元输出特殊化
printf("%d",-pans[pans[0]]);
else
printf("%d",pans[pans[0]]);
for(ca=pans[0]-1;ca>=1;ca--)
{
if(pans[ca]>=1000)//4位全满
printf("%d",pans[ca]);
else
{
//if(ca {
temp=pans[ca];
for(cb=0;temp>0;cb++)//输出数字中间的0
temp/=10;
for(;cb printf("0");
}
if(pans[ca]!=0) printf("%d",pans[ca]);//当此单元为0时，0已经输出完毕，就不用再输出了
}
}
printf("\n");
return 0;
}
short WhoBig(short *pnum1,short *pnum2)
{
short ca;
if(pnum1[0]>pnum2[0])
return 1;
else
if(pnum1[0] return -1;
for(ca=pnum1[0];ca>=1;ca--)
{
if(pnum1[ca]==pnum2[ca])
continue;
if(pnum1[ca]>pnum2[ca])
return 1;
else
return -1;
}
return 0;
}
int main()
{
short n,ca,big;
char strleft[TMAX][NUMLONG+1],strright[TMAX][NUMLONG+1];
short numleft[TMAX][NUMLONG/DIV]={0},numright[TMAX][NUMLONG/DIV]={0};
short ans[TMAX][NUMLONG/DIV]={0};
scanf("%d",&n);
ca=0;
while(ca {
ChangetoNum(strleft[ca],numleft[ca],ca);//正正为3，正负为4，负正为1，负负为2
if(deadd[ca]==0)
deadd[ca]=3;
ChangetoNum(strright[ca],numright[ca],ca);
ca++;
}//while输入数据
for(ca=0;ca {
switch(deadd[ca])
{
case 3:
Calculate(numleft[ca],numright[ca],ans[ca]);
flag=0;
break;
case 2:
Calculate(numleft[ca],numright[ca],ans[ca]);
flag=1;
break;
case 4:
case 1:
big=WhoBig(numleft[ca],numright[ca]);
if(big==0)
{
printf("Case %d:\n",ca+1);
printf("%s + %s = 0\n",strleft[ca],strright[ca]);
continue;
}
else
if(big==1)//左边的数大于右边的数
{
Calculate2(numleft[ca],numright[ca],ans[ca]);
if(deadd[ca]==4)//正负
flag=0;
else
if(deadd[ca]==1)//负正
flag=1;
}
else
if(big==-1)//右边的数大于左边的数
{
Calculate2(numright[ca],numleft[ca],ans[ca]);
if(deadd[ca]==4)//正负
flag=1;
else
if(deadd[ca]==1)//负正
flag=0;
}
break;
}
PrintAns(strleft[ca],strright[ca],ans[ca],ca+1);
if(ca!=n-1)
printf("\n");
}//for处理数据
return 0;
}
