两个无穷大的数相加的问题
这是原题, 哪位高手能知道一下啊?
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11377 Accepted Submission(s): 1904
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
Author
Ignatius.L
[解决办法]
#include<stdio.h>#include<string.h>int main(){ char str1[1000],str2[1000]; int a[1000],b[1000],c[1001]; int N; int m,n,i,j,s,x,k,h,t; int flag=0; int len1,len2,len; scanf("%d",&N); for(i=1;i<=N;++i) { scanf("%s %s",&str1,&str2); printf("Case "); printf("%d",i); printf(":\n"); printf("%s + %s = ",str1,str2); len1=strlen(str1); len2=strlen(str2); for(s=len1-1,m=0;s>=0;s--) { a[m]=str1[s]-'0'; m++; } for(j=len2-1,n=0;j>=0;j--) { b[n]=str2[j]-'0'; n++; } for(h=len1;h<1000;h++) a[h]=0; for(t=len2;t<1000;t++) b[t]=0; if(len1>len2) len=len1; else if(len1=len2) len=len1; else len=len2; for(s=0;s<len;s++) { x=a[s]+b[s]+flag; if(x>=10) { c[s]=x%10; flag=1; } else { c[s]=x; flag=0; } } if(flag==1) printf("1"); for(k=len-1;k>=0;k--) printf("%d",c[k]); printf("\n"); if (i<N) printf("\n"); } return 0;}
[解决办法]
#include <stdio.h> #include <string.h> int main(){ char a[1005],b[1005]; int kase,ase=1;; scanf("%d",&kase); while (kase--){ // while (scanf("%s",a)!=EOF){ scanf ("%s%s",a,b); if (ase>1) printf("\n"); printf("Case %d:\n",ase++); int la,lb; la=strlen(a); lb=strlen(b); printf("%s + %s = ",a,b); char c[2020]; int i,j,k=0,jw=0; for (i=la-1,j=lb-1;i>=0&&j>=0;i--,j--){ c[k++]=(a[i]-'0'+b[j]-'0'+jw)%10+'0'; jw=(a[i]-'0'+b[j]-'0'+jw)/10; } if (i==-1){ for (j;j>=0;j--){ c[k++]=(b[j]-'0'+jw)%10+'0'; jw=(b[j]-'0'+jw)/10; } } else { for (i;i>=0;i--){ c[k++]=(a[i]-'0'+jw)%10+'0'; jw=(a[i]-'0'+jw)/10; } } if(jw!=0) printf ("%d",jw); for (k-=1;k>=0;k--) printf ("%d",c[k]-'0'); printf("\n"); } return 0 ; }