hdu 1002 高精度加法
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A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 254093 Accepted Submission(s): 49006
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
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
Author
Ignatius.L
Recommend
思路:①倒置对齐②逐位相加③逐位检查是否大于9,若大于9,则施行进位④重新倒置输出
#include<stdio.h>#include<string.h>#define min(x,y) (x<=y?x:y)#define maxn 1001int main(){int a,last,len1,len2,x[maxn],y[maxn],k,i,j;char b[maxn],c[maxn];scanf("%d",&a);for(j=1;j<=a;j++){x[0]=0; y[0]=0;scanf("%s%s",b,c);len1=strlen(b);len2=strlen(c);for(i=len1-1;i>=0;i--){x[0]++;x[x[0]]=b[i]-'0';}for(i=len2-1;i>=0;i--){y[0]++;y[y[0]]=c[i]-'0';}last=0;k=min(x[0],y[0]);for(i=1;i<=k;i++){x[i]+=y[i]+last;last=x[i]/10;x[i]=x[i]%10;}if(len1>k){for(i=k+1;i<=len1;i++)if(last>0){x[i]+=last;last=x[i]/10;x[i]%=10;}}if(len2>k){for(x[0]=y[0],i=k+1;i<=len2;i++){x[i]=y[i]+last;last=x[i]/10;x[i]%=10;}} printf("Case %d:\n",j); printf("%s + %s = ",b,c); for(i=x[0];i>=1;i--) { printf("%d",x[i]);}if(j!=a){printf("\n\n");}else{printf("\n");}}return 0;}
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