A+B Problem II
时间限制:3000 ms | 内存限制:65535 KB
难度:3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
A,B must be positive.
输入
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.
输出
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.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
*/
#include<stdio.h> #include<string.h> int main(){ int n; int len1,len2,i,j; int t; int k=0; scanf("%d",&n); while(n--){ char a1[1100],a2[1100];// 开始时,数组定义过小,导致超时 int a3[1100],a4[1100]; int b2[1100]; memset(a3,0,sizeof(a3)); memset(a4,0,sizeof(a4)); memset(b2,0,sizeof(b2)); scanf("%s%s",a1,a2); k++; printf("Case %d: ",k); printf("%s",a1); printf(" + "); printf("%s",a2); printf(" = "); len1=strlen(a1); len2=strlen(a2); //将两个字符数倒序,底位在前,高位在后 for(i=0;i<len1;i++){ a3[i]=a1[i]-'0'; } for(j=0;j<len2;j++){ a4[j]=a2[j]-'0'; } for(i=0;i<(len1/2);i++) { t=a3[i]; a3[i]=a3[len1-i-1]; a3[len1-i-1]=t; } for(j=0;j<(len2/2);j++){ t=a4[j]; a4[j]=a4[len2-j-1]; a4[len2-j-1]=t; } if(len1<len2) len1=len2; for(i=0;i<=len1;i++) // 进位,相加大于9时,向下一位进1,并且将本位去十 { b2[i]=(a3[i])+(a4[i]); if(b2[i]>9){ a3[i+1]++; b2[i]-=10; } } int i=len1+2; while(b2[i]==0)i--;//将高位的零去掉,并且倒序一次输出 for(;i>0;i--){ printf("%d",b2[i]); } printf("%d ",b2[0]); if(n) printf(" "); } return 0; }