題目
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
代碼
由於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.
所以必須用字符串實現大數加法
大數加法
-
判斷是否爲特殊情況
1.1a,b字符串皆爲空，則輸出0
1.2a字符串爲空，則輸出b
1.3b字符串爲空，則輸出a -
翻轉a,b字符串
因爲大數加法是從低位往高位加，而字符串是從高位往低位寫入的，所以需要對a,b字符串進行翻轉，這裏運用的是reverse函數
頭文件要加algorithm
reverse(N.begin(), N.end()) -
總是讓a作爲較短的那個字符串,而b作爲目標字符串（每一位的加法只需要循環到lena即可）
這裏寫了兩個swap函數來交換字符串和表示字符串長度的變量
void swapstring(string &a, string &b)
{
string t = a;
a = b;
b = t;
return;
}
void swap(int &a, int &b){
int t = a;
a = b;
b = t;
return;
}
-
設置進位標誌flag
-
a,b都有數的每一位的加法計算
5.1用isdigit判斷加後的和數是否仍爲0~9的數字
5.2若當前進位符爲true則+1且flag = false，回溯爲不進位
5.3若不爲數字，則flag = true進位，並且b[i] -= 10 -
a被用完了若還有進位則靠這一位進位還要與剩餘的b字符串的高位數進行加法運算
循環直到進位符爲false,表示不再進位了6.1如果長度相等，b字符串最高位添一位1，flag = false
6.2否則在走到的當前位上+1
~6.2.1加後如果不是數字，進位符flag = true，且b[lena] -= 10
~6.2.2如果是數字，flag = false -
翻轉目標字符串b，返回b
string SUM(string a, string b){
if (a == ""&&b == "")
return "0";
if (a == "")
return b;
if (b == "")
return a;
/*特殊情況*/
int lena = a.length(),lenb=b.length();
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
bool flag = false;
if (lena > lenb){
swap(lena, lenb);
swapstring(a, b);
}
for (int i = 0; i < lena; i++){
b[i] += a[i] - '0';
if (flag){
b[i] += 1;
flag = false;
}
if (!isdigit(b[i])){
flag = true;
b[i] -= 10;
}
}
while(flag){//較短的最高位進位不知道要進幾次位
if (lena == lenb){
b += '1';
flag = false;
}
else{
b[lena] += 1;
if (!isdigit(b[lena])){
flag = true;
b[lena] -= 10;
}
else
flag = false;
}
lena++;
}
reverse(b.begin(), b.end());
return b;
}
AC代碼
#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<stdlib.h>
using namespace std;
void swapstring(string &a, string &b)
{
string t = a;
a = b;
b = t;
return;
}
void swap(int &a, int &b){
int t = a;
a = b;
b = t;
return;
}
string SUM(string a, string b){
if (a == ""&&b == "")
return "0";
if (a == "")
return b;
if (b == "")
return a;
/*特殊情況*/
int lena = a.length(),lenb=b.length();
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
bool flag = false;
if (lena > lenb){
swap(lena, lenb);
swapstring(a, b);
}
for (int i = 0; i < lena; i++){
b[i] += a[i] - '0';
if (flag){
b[i] += 1;
flag = false;
}
if (!isdigit(b[i])){
flag = true;
b[i] -= 10;
}
}
while(flag){//較短的最高位進位不知道要進幾次位
if (lena == lenb){
b += '1';
flag = false;
}
else{
b[lena] += 1;
if (!isdigit(b[lena])){
flag = true;
b[lena] -= 10;
}
else
flag = false;
}
lena++;
}
reverse(b.begin(), b.end());
return b;
}
int main()
{
int t;
scanf("%d", &t);
for(int i=1;i<=t;i++)
{
string a, b;
cin >> a >> b;
cout << "Case " << i << ":" << endl;
cout << a << " + " << b << " = " << SUM(a, b) << endl;
if (i != t)
cout << endl;
}
system("pause");
return 0;
}