1 条题解
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0
C :
#include<stdio.h> int main() { int t; scanf("%d",&t); while(t--) { int x,y,k,ans; scanf("%d%d%d",&x,&y,&k); if(y<=x) ans=x; else if(k>=y-x) ans=y; else ans=y+y-x-k; printf("%d\n",ans); } return 0; }C++ :
#include <bits/stdc++.h> #define int long long using namespace std; const int N=1e6+10; typedef pair<int,int> PII; typedef priority_queue<int , vector<int>, greater<int>> minqueue; //从小到大 queue typedef priority_queue<int, vector<int>, less<int>> maxqueue; //从大到小 queue int gcd(int a, int b){ //最大公因数 return b ? gcd(b, a % b) : a; } int lcm(int a, int b){ //最小公倍数 return a * b / __gcd(a, b); } int qmi(int base, int power, int p) //快速幂求余 { int result = 1; while (power > 0) { if (power & 1) result = result * base % p; base = base * base % p ; power >>= 1; } return result % p; } void solve() { int x,y,k; cin>>x>>y>>k; if(x>=y){ cout<<x<<'\n'; } else{ if(x+k>=y) cout<<y<<'\n'; else cout<<y+y-(x+k)<<'\n'; } } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; t=1; cin>>t; while(t--) { solve(); } return 0; }Java :
import java.util.Scanner; public class Main { public static void main(String[] args){ Scanner sc=new Scanner(System.in); int t= sc.nextInt(); for (int i = 0; i <t; i++) { int x= sc.nextInt(); int y= sc.nextInt(); int k= sc.nextInt(); int r; if(x>y){ r=x; System.out.println(r); } else if(x<y&&(x+k)>=y){ r=y; System.out.println(r); } else if (x<y&&(x+k)<y) { r=2*y-x-k; System.out.println(r); } } } }
- 1
信息
- ID
- 722
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 10
- 标签
- 提交数
- 8
- 已通过
- 2
- 上传者