Greedy Takahashi——UPC

题目描述

Takahashi has A cookies, and Aoki has B cookies. Takahashi will do the following action K times:
·If Takahashi has one or more cookies, eat one of his cookies.
·Otherwise, if Aoki has one or more cookies, eat one of Aoki’s cookies.
·If they both have no cookies, do nothing.
In the end, how many cookies will Takahashi and Aoki have, respectively?

Constraints
·0≤A≤10¹²
·0≤B≤10¹²
·0≤K≤10¹²
·All values in input are integers.

输入

Input is given from Standard Input in the following format:

A B K

输出

Print the numbers of Takahashi’s and Aoki’s cookies after K actions.

样例输入

【样例1】
2 3 3
【样例2】
500000000000 500000000000 1000000000000

样例输出

【样例1】
0 2
【样例2】
0 0

提示

样例1解释
Takahashi will do the following:
·He has two cookies, so he eats one of them.
·Now he has one cookie left, and he eats it.
·Now he has no cookies left, but Aoki has three, so Takahashi eats one of them.
Thus, in the end, Takahashi will have 0 cookies, and Aoki will have 2.
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很水，直接放代码

#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
start{
    ll a,b,k;
    cin>>a>>b>>k;
    if(k<=a) cout<<a-k<<" "<<b<<endl;
    else if(k<=a+b) cout<<0<<" "<<b+a-k<<endl;
    else cout<<0<<" "<<0<<endl;
    end;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:1 ms
    Memory:2024 kb
****************************************************************/