CodeForces - 907A Masha and Bears
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size a can climb into some car with size b if and only if a ≤ b, he or she likes it if and only if he can climb into this car and 2a ≥ b.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
You are given four integers V1, V2, V3, Vm(1 ≤ Vi ≤ 100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that V1 > V2 > V3.
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
50 30 10 10
50
30
10
100 50 10 21
-1
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
分析
水题,但有个坑点,就是Masha只like最小的车,就是说2*Vm<v2。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<algorithm> #include<cstring> #include<queue> using namespace std; typedef long long LL; const int maxn = 1e4+5; const int mod = 772002+233; typedef pair<int,int> pii; #define X first #define Y second #define pb push_back #define mp make_pair #define ms(a,b) memset(a,b,sizeof(a)) int main(){ // freopen("in.txt","r",stdin); int v1,v2,v3,vm; scanf("%d%d%d%d",&v1,&v2,&v3,&vm); if(vm>2*v3||v3>2*vm||v2<=vm){ cout<<-1<<endl; }else{ cout<<2*v1<<endl<<2*v2<<endl<<max(v3,vm)<<endl; } return 0; }