5 条题解

  • 3
    @ 2023-10-1 22:55:01
    #include <stdio.h>
    int main(){
        int N;
        scanf("%d",&N);
           N=N/10;
        switch(N){
            case 9 ... 10 :
            printf("Great");
            break;
            case 7 ... 8 :
            printf("Good");
            break;
               case 6:
            printf("Average");
            break;
               case 0 ... 5 :
            printf("Poor");
            break;
              }
    
        return 0;
    }
    
    
    • 1
      @ 2024-12-11 19:26:27

      #include<stdio.h> int main() { double n; scanf("%lf",&n); int x=int(n/10); switch(x) { case 0: case 1: case 2: case 3: case 4: case 5: printf("Poor"); break; case 6: printf("Average"); break; case 7: case 8: printf("Good"); break; case 9: case 10: printf("Great"); break; } return 0; }

      • 0
        @ 2026-4-25 21:22:43

        都是反骨仔 #include <stdio.h>

        int main() { int score;

        scanf("%d", &score);
        
        
        if (score >= 90) {
            printf("Great\n");
        } else if (score >= 70) {
            
            printf("Good\n");
        } else if (score >= 60) {
            
            printf("Average\n");
        } else {
            
            printf("Poor\n");
        }
        
        return 0;
        

        }

        • 0
          @ 2023-10-6 11:49:04
          #include<stdio.h>
          int main(){
              int n;
              scanf("%d",&n);
              n/=10;
              switch(n){
                  case 10 :printf("Great");break;
                  case 9 :printf("Great");break;
                  case 8 :printf("Good");break;
                  case 7 :printf("Good");break;
                  case 6 :printf("Average");break;
                  default :printf("Poor");break;
              }
              return 0;
          }
          
          • 0
            @ 2023-8-13 22:51:59

            #include<stdio.h> int main() {int n; scanf("%d",&n); if(n >= 90 && n <= 100) {printf("Great"); } else if(n >= 70 && n <= 89) {printf("Good"); } else if(n >= 60 && n <= 69) {printf("Average"); } else if(n >= 0 && n <= 59) {printf("Poor"); } return 0; }

            • 1

            信息

            ID
            67
            时间
            1000ms
            内存
            128MiB
            难度
            2
            标签
            (无)
            递交数
            1725
            已通过
            1001
            上传者