A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 317574    Accepted Submission(s): 61700 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   #include <stdio.h> #include <string.h> #define MAX 1005 using namespace std; int a[MAX], b[MAX]; int sum[MAX]; void init () { memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); memset(sum, 0, sizeof(sum)); } int bigNumAdd ( char ca[], char cb[] ) { int i; int len_ca = strlen(ca); int len_cb = strlen(cb); /*进制数值*/ int carry = 0; /*将数值进行整数化，然后再进行反转*/ for ( i = 0;i < len_ca; i++ ) { a[i] = ca[len_ca-i-1]-'0'; } for ( i = 0;i < len_cb; i++ ) { b[i] = cb[len_cb-i-1]-'0'; } /*找到最长的数值的长度*/ int len_result = (len_ca > len_cb ? len_ca : len_cb); /*代码中最重要的部分*/ for ( i = 0;i < len_result; i++ ) { sum[i] = (a[i]+b[i]+carry)%10; carry = (a[i]+b[i]+carry)/10; } /*如果最后进制不为0*/ if ( carry ) { sum[i] = carry; len_result++; } /*返回和数的长度*/ return len_result; } void NormalNum( char c[] ) { int i, j; int len = strlen(c); i = 0; while ( c[i] == '0'&&i != len-1 ) { i++; } j = 0; while ( i <= len ) { c[j++] = c[i++]; } } int main() { int t; int i, j; char ca[MAX], cb[MAX]; scanf ( "%d", &t ); for ( i = 1;i <= t; i++ ) { /*初始化*/ init(); scanf ( "%s %s", ca, cb ); /*数值的规范化，本题这步可以忽略*/ NormalNum (ca); NormalNum (cb); int len_result = bigNumAdd (ca, cb); printf ( "Case %d:\n%s + %s = ", i, ca, cb ); for ( j = len_result-1;j >= 0; j-- ) { printf ( "%d", sum[j] ); } printf ( "\n" ); if ( i != t) printf ( "\n" ); } return 0; } 简化的代码（就简化了一点） #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #define MAX 1005 using namespace std; char ca[MAX], cb[MAX]; int a[MAX], b[MAX]; int sum[MAX]; int Max; void init () { int i; int len_ca = strlen(ca); int len_cb = strlen(cb); memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); memset(sum, 0, sizeof(sum)); Max = max(len_ca, len_cb)+2; for ( i = 0;i < len_ca; i++ ) { a[i] = ca[len_ca-i-1]-'0'; } for ( i = 0;i < len_cb; i++ ) { b[i] = cb[len_cb-i-1]-'0'; } } void bigNumAdd () { int i; int carry = 0; for ( i = 0; i < Max; i++ ) { sum[i] = (a[i]+b[i]+carry)%10; carry = (a[i]+b[i]+carry)/10; } } int main() { int t; int i, j; scanf ( "%d", &t ); for ( j = 1;j <= t; j++ ) { scanf ( "%s %s", ca, cb ); init(); bigNumAdd(); printf ( "Case %d:\n%s + %s = ", j, ca, cb ); for ( i = Max; i >= 0; i-- ) { if(sum[i] != 0) break; } for ( i;i >= 0; i-- ) { printf ( "%d", sum[i] ); } printf ( "\n" ); if ( j!= t) { printf ( "\n" ); } } return 0; } 代码菜鸟，如有错误，请多包涵！！！

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