HDOJ -- 1002 大数A+B
发布时间：2021-02-07 02:45:42 所属栏目：大数据 来源：网络整理
导读：A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Problem Description I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B. ? Input Th
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A + B Problem IITime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Problem Description I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.? Input 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. ? Output 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. Output a blank line between two test cases. ? Sample Input 2 1 2 112233445566778899 998877665544332211? Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 2222222222222221110?题解：对于此题做法有两种：其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，方便运算。 下面用法一，代码如下: #include<stdio.h>
#include<string.h>
int main()
{
int n,i,len1,len2,j,k,pi,t,flag;
char str1[1010],str2[1010];
scanf("%d",&n);
for(i=1;i<=n;i++)
{
int a[1200]={0};
flag=0;
printf("Case %d:n",i);
scanf("%s%s",str1,str2);//以字符串形式读入
len1=strlen(str1);
len2=strlen(str2);
printf("%s + %s = ",str2);
j=len1-1;
k=len2-1;
pi=0;
while(j>=0&&k>=0)//开始相加
{
if(a[pi]+(str1[j]-'0')+(str2[k]-'0')>=10)//相加后大于10的
{
a[pi]=a[pi]+(str1[j]-'0')+(str2[k]-'0')-10;
a[pi+1]++;
}
else
a[pi]=a[pi]+(str1[j]-'0')+(str2[k]-'0');
pi++;
k--;
j--;
}
if(j>=0)
{
for(t=j;t>=0;t--)
{
a[pi]=a[pi]+(str1[t]-'0');
pi++;
}
}
else if(k>=0)
{
for(t=k;t>=0;t--)
{
a[pi]=a[pi]+str2[t]-'0';
pi++;
}
}
else if(a[pi]!=0)//对于位数相同2个数加后最高位大于10的
pi++;
for(t=pi-1;t>=0;t--)
{
if(a[t]==0&&flag==0)//处理一次啊前导0，估计属于无用的一步
continue;
else
{
flag=1;
printf("%d",a[t]);
}
}
printf("n");
if(i!=n)//对于2组之间加空行的情况
printf("n");
}
return 0;
}
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