2562.txt

来自「北大ACM题目例程 详细的解答过程 程序实现 算法分析」· 文本 代码 · 共 32 行

main()
{char a[110],b[110];
int i,at,bt,temp;
while(1)
{scanf("%s",a);scanf("%s",b);
if(a[0]=='0'&&b[0]=='0')exit(0);

at=0;bt=0;i=0;temp=0;

while(a[at+1]!=0)at++;
while(b[bt+1]!=0)bt++;

while(at>=0&&bt>=0)
{if(a[at]-'0'+b[bt]-'0'+temp>=10){i++;temp=1;}else temp=0;
at--;bt--;}

while(at>=0)if(temp+a[at]-'0'>=10){i++;temp=1;at--;}else break;

while(bt>=0)if(temp+b[bt]-'0'>=10){i++;temp=1;bt--;}else break;

if(i==1)printf("%d carry operation.\n",i);


if(i>1)printf("%d carry operations.\n",i);
if(i==0)printf("No carry operation.\n");
}
}