hdoj1002A + B Problem II
2015-09-08 21:42
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A + B Problem II
[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 266282 Accepted Submission(s): 51573
[/b]
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 = 1111111111111111110
一看就是大数!!
代码:
#include<stdio.h> #include<string.h> char al[1010],bl[1010]; int a[1010],b[1010]; int main() { int i,j,n; scanf("%d",&n); getchar(); int k=1; while(n--) { scanf("%s",al);//存为字符串 scanf("%s",bl); int l1=strlen(al); int l2=strlen(bl); memset(a,0,sizeof(a));//初始化 memset(b,0,sizeof(b)); printf("Case %d:\n",k++); for(j=0,i=l1-1;i>=0;i--)//转化为数字 { a[j]=al[i]-'0'; j++; } for(j=0,i=l2-1;i>=0;i--) { b[j]=bl[i]-'0'; j++; } for(i=0;i<1009;i++)//位位相加 { a[i]=a[i]+b[i]; if(a[i]>=10) { a[i]-=10; a[i+1]++;//如果大于十下一位加一 } } printf("%s + %s = ",al,bl); for(i=1009;i>=0;i--) { if(a[i]!=0)//去掉前导零 break; } if(i>=0) for(;i>=0;i--) { printf("%d",a[i]); } printf("\n"); if(n!=0) printf("\n"); } }
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