A+B Problem II
时间限制:
3000?ms ?|? 内存限制:
65535?KB
难度:
3
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描述
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I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.
A,B must be positive.
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输入
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The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow,each line consists of two positive integers,A and B. Notice that the integers are very large,that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
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输出
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For each test case,you should output two lines. The first line is "Case #:",# means the number of the test case. The second line is the an equation "A + B = Sum",Sum means the result of A + B. Note there are some spaces int the equation.
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样例输入
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2
1 2
112233445566778899 998877665544332211
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样例输出
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Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110
解题方法:
? ? ? ? ? ? ?先将字符串反转存入整形数组,整形数组按位相加,若相加后大于十则前一位加一。相加结束后,
? ? ?倒叙逐个输出整形数组的每一位数。
代码如下:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int max(int x,int y)
{
int z;
z=x>y ? x:y;
return z;//比较两个数的大小,将较大的数赋值给返回值
}
int main()
{
int i,n,t=0,z,lena,lenb,num1[1100],num2[1100];
char a[1100],b[1100];
scanf("%d",&n);
while(n--)
{
memset(num1,sizeof(num1));
memset(num2,sizeof(num2));//int型数组清零
scanf("%s%s",&a,&b);
lena=strlen(a);
lenb=strlen(b);
for(i=0;i<lena;i++)
num1[lena-1-i]=a[i]-'0';
for(i=0;i<lenb;i++)
num2[lenb-1-i]=b[i]-'0';
z=max(lena,lenb);
for(i=0;i<z;i++)//此处i值的范围就是两个字符串长度的最大者
{
num1[i]=num1[i]+num2[i];
if(num1[i]>9)//处理相加大于十的情况
{
num1[i]=num1[i]-10;
num1[i+1]++;
}
}
t++;
if(num1[z]==0)// 判断数组第z位是否为零
{
printf("Case %d:n",t);
printf("%s + %s = ",a,b);
for(i=z-1;i>=0;i--)
printf("%d",num1[i]);
printf("n");
}
else
{
printf("Case %d:n",b);
for(i=z;i>=0;i--)
printf("%d",num1[i]);
printf("n");
}
}
return 0;
}
(编辑:李大同)
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