1 solutions
-
0
C :
#include<stdio.h> main() { int x,y,z; scanf("%d %d %d %d",&x,&y,&z); double a = (x + y + z) / 3.0; if(a >= 90) { int n = 0; //统计在90分以上的科目数量 if(x >= 90) { n++; } if(y >= 90) { n++; } if(z >= 90) { n++; } if(n == 3) { printf("beijing"); } else if(n == 2) { printf("xiamen"); } else { printf("guilin"); } } else if(a >= 85) { if(x >= 85 && y >= 85 && z >= 85) { printf("shanghai"); } else { printf("nanjing"); } } }
C++ :
#include <bits/stdc++.h> using namespace std; int main(){ int x,y,z,c = 0; cin>>x>>y>>z; double v = (x + y + z) / 3.0; if(v >= 90){ if(x >= 90) c++; if(y >= 90) c++; if(z >= 90) c++; if(c == 3) cout<<"beijing"; else if(c == 2) cout<<"xiamen"; else cout<<"guilin"; }else if(v >= 85){ if(x >= 85) c++; if(y >= 85) c++; if(z >= 85) c++; if(c == 3) cout<<"shanghai"; else cout<<"nanjing"; } }
Java :
import java.util.Scanner; public class Main { public static void main(String[] Args) { Scanner sc = new Scanner(System.in); int a = sc.nextInt(); int b = sc.nextInt(); int c = sc.nextInt(); int d = (a+b+c)/3; int i = 0; int j = 0; if(a>=90) ++i; else if(a>=85) ++j; if(b>=90) ++i; else if(b>=85) ++j; if(c>=90) ++i; else if(c>=85) ++j; if(d>=90){ if(i == 3) System.out.println("beijing"); if(i == 2) System.out.println("xiamen"); if(i == 1) System.out.println("guilin"); } else if(d>=85){ if(j == 3) System.out.println("shanghai"); else System.out.println("nanjing"); } } }
Python :
# 输入字符串切割 s = input().split() x = int(s[0]) y = int(s[1]) z = int(s[2]) # 平均分v v = (x + y + z) / 3 # 判断是远途还是周边 if v >= 90: #远途 if x >= 90 and y >= 90 and z >= 90: print("beijing") elif (x >= 90 and y >= 90) or (x >= 90 and z >= 90) or (y >= 90 and z >= 90): print("xiamen") else: print("guilin") elif 85 <= v < 90: #周边 if x>=85 and y>=85 and z>=85: print("shanghai") else: print("nanjing")
- 1
Information
- ID
- 10257
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By