A + B Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 1538 Accepted Submission(s): 609 |
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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.
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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. |
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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. |
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Sample Input
2 1 2 112233445566778899 998877665544332211 |
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Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 |
简单的大数据运算。。。直接用的模版
ac代码g++：
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <iostream>
#define rep(i,s,e) for(int i= (s);i<=(e);++i)
using namespace std;
const int maxn = 1010;
struct bign{
int len, s[maxn];
bign() {
memset(s, 0, sizeof(s));
len = 1;
}
bign(int num) {
*this = num;
}
bign(const char* num) {
*this = num;
}
bign operator = (int num) {
char s[maxn];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char* num) {
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
string str() const {
string res = "";
for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
if(res == "") res = "0";
return res;
}
bign operator + (const bign& b) const{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
void clean() {
while(len > 1 && !s[len-1]) len--;
}
bign operator * (const bign& b) {
bign c; c.len = len + b.len;
for(int i = 0; i < len; i++)
for(int j = 0; j < b.len; j++)
c.s[i+j] += s[i] * b.s[j];
for(int i = 0; i < c.len-1; i++){
c.s[i+1] += c.s[i] / 10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator - (const bign& b) {
bign c; c.len = 0;
for(int i = 0, g = 0; i < len; i++) {
int x = s[i] - g;
if(i < b.len) x -= b.s[i];
if(x >= 0) g = 0;
else {
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}
bool operator < (const bign& b) const{
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--)
if(s[i] != b.s[i]) return s[i] < b.s[i];
return false;
}
bool operator > (const bign& b) const{
return b < *this;
}
bool operator <= (const bign& b) {
return !(b > *this);
}
bool operator == (const bign& b) {
return !(b < *this) && !(*this < b);
}
bign operator += (const bign& b) {
*this = *this + b;
return *this;
}
};
istream& operator >> (istream &in, bign& x) {
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream &out, const bign& x) {
out << x.str();
return out;
}
int main() {
//freopen("in.txt","r",stdin);
bign a,b;
int kcase = 1;
int t;
scanf("%d",&t);
int first = 0;
while(t--)
{
cin>>a>>b;
if(first)
printf("\n");
first = 1;
printf("Case %d:\n",kcase++);
cout <<a<<" + "<<b<<" = "<<a+b << endl;
// cout<<endl;
}
return 0;
}