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
#include"stdio.h" #include"string.h" int main() { const int maxn=1e5; char a[maxn+1],b[maxn+1]; int c[maxn+1],j,k,g,h; int n,Al,Bl,sum; scanf("%d",&n); for (int i=1;i<=n;i++) { scanf("%s %s",a,b); getchar(); Al=strlen(a); Bl=strlen(b); if(i!=1) printf("\n"); printf("Case %d:\n",i); printf("%s",a); printf(" + "); printf("%s",b); printf(" = "); if (Al>=Bl) { for (j=Bl-1;j>=0;j--) { b[j+Al-Bl]=b[j]; } for (k=0;k<Al-Bl;k++) { b[k]='0'; } for (g=Al-1;g>0;g--) { sum=a[g]+b[g]-96; if (sum>=10) { c[g]=sum-10; a[g-1]=a[g-1]+1; } else c[g]=sum; } if (a[0]+b[0]-96>=10) { c[0]=a[0]+b[0]-106; printf("1"); for (h=0;h<Al;h++) { printf("%d",c[h]); } } else { c[0]=a[0]+b[0]-96; for (h=0;h<Al;h++) { printf("%d",c[h]); } } printf("\n"); } else { for (j=Al-1;j>=0;j--) { a[j+Bl-Al]=a[j]; } for (k=0;k<Bl-Al;k++) { a[k]='0'; } for (g=Bl-1;g>0;g--) { sum=a[g]+b[g]-96; if (sum>=10) { c[g]=sum-10; b[g-1]=b[g-1]+1; } else c[g]=sum; } if (a[0]+b[0]-96>=10) { c[0]=a[0]+b[0]-106; printf("1"); for (h=0;h<Al;h++) { printf("%d",c[h]); } } else { c[0]=a[0]+b[0]-96; for (h=0;h<Bl;h++) { printf("%d",c[h]); } } printf("\n"); } } return 0; }or
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main () { char a[8000],b[8000]; int na[8000],nb[8000],sum[8000],pre,flag=1; int t; scanf("%d",&t); while (t--) { memset(sum,0,sizeof(sum)); memset(na,0,sizeof(na)); memset(nb,0,sizeof(nb)); scanf("%s%s",a,b); pre=0; int lena=strlen(a); int lenb=strlen(b); for (int i=0; i<lena; i++) na[lena-1-i]=a[i]-'0'; for (int j=0; j<lenb; j++) nb[lenb-1-j]=b[j]-'0'; int lenx=lena>lenb?lena:lenb; for (int k=0; k<lenx; k++) { sum[k]=na[k]+nb[k]+pre/10; pre=sum[k]; } while (pre>9) { sum[lenx]=pre/10%10; lenx++; pre/=10; } printf ("Case %d:\n",flag++); printf ("%s + %s = ",a,b); for (int i=lenx-1; i>=0; i--) { printf ("%d",sum[i]%10); } printf ("\n"); if (t) printf ("\n"); } return 0; }大数加法
