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
第六次尝试的代码(/”≡ _ ≡)/~┴┴
#include
#include
void accumlate(char* a, char* b, char* c, int maxl);
void reverse(char* a, int len);
int toNum(char a);
char toChar(int a);
char num1[2000];
char num2[2000];
char answer[2000];
int main(void) {
int times, i;
scanf("%d", ×);
for(i=0; imemset(num1,0,sizeof(num1));
memset(num2,0,sizeof(num1));
memset(answer,0,sizeof(answer));
scanf("%s", num1);
scanf("%s", num2);
int len1 = strlen(num1);
int len2 = strlen(num2);
int maxl = len1>len2?len1:len2;
printf("Case %d:\n",i+1);
printf("%s + %s = ", num1, num2);
reverse(num1, len1);
reverse(num2, len2);
accumlate(num1,num2,answer,maxl);
printf("%s", answer);
//这里的格式, emmmmmm
i==times-1?printf("\n"):printf("\n\n");
}
return 0;
}
void reverse(char* a, int len) {
char temp;
int i,j;
for(i=0,j=len-1; i<=j; i++,j--) {
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
int toNum(char a) {
if(a=='\0')
return 0;
return (int)a - 48;
}
char toChar(int a) {
return (char)(a+48);
}
void accumlate(char* a, char* b, char* c, int maxl) {
int i, carry=0, temp=0;
for(i=0; i1 ;i++) {
if(carry==0 && i==maxl) break;
temp = toNum(a[i])+toNum(b[i])+carry;
c[i] = toChar(temp%10);
carry = temp/10;
}
reverse(c, strlen(c));
}
reverse
的时候没有弄清楚,以后要记住,奇偶两者情况的判定都是i <= j
就是头不能比尾小。