5 条题解

  • 0
    @ 2025-11-2 2:24:14
    #include <stdio.h>
    int main() {
        int N;
        scanf("%d", &N);
        long long n,i=0;
        while(N--){
            i++;
            scanf("%lld", &n);
            long long a = n * (n + 1) * (n + 2) * (n + 3) / 8;
            printf("%lld %lld %lld\n", i, n, a);
        }
    }
    
    • 0
      @ 2025-10-30 21:10:35
      #include<stdio.h>
      int add(int k){
      	int sum = 0;
      	for(int i = 1; i <= k; i++){
      		int s = 0;
      		for(int j = 1; j <= i + 1; j++){
      			s += j;
      		}
      		sum += i * s;
      	}
      	return sum;
      }
      int main(){
      	int N,num = 1;
      	scanf("%d",&N);
      	while(N--){
      		int n;
      		scanf("%d",&n);
      		printf("%d %d %d\n",num,n,add(n));
      		num++;
      	}
      	return 0;
      }
      
      • 0
        @ 2025-10-5 16:25:16

        #include<stdio.h>

        int main(){

        int t,n,a,x=1;

        scanf("%d",&t);

        while(t--){

        scanf("%d",&n);

        int sum=0;

        for(int i=1;i<=n;i++){

        a=i+1;

        sum=sum+i*((a*a+a)/2);

        }

        printf("%d %d %d\n",x,n,sum);

        x++;

        }

        return 0;

        }

        • 0
          @ 2024-11-1 17:09:43

          #define _CRT_SECURE_NO_WARNINGS #include<stdio.h> int arr[10001];//存储前n个数的和 void nat() { int num = 0; int j = 1; for (int i = 1; i < 10001; i++) { num += i; arr[j++] = num;//j表示前j个自然数的和 } }

          int add(int a) { nat(); int sum = 0; for (int i = 1; i < a+1 ; i++) { sum += i * arr[i + 1]; } return sum; }

          int main() { int n; int hao = 0; scanf("%d", &n); while (n--) { hao++; int a; scanf("%d", &a); printf("%d %d %d\n",hao,a, add(a)); } return 0; }

          • 0
            @ 2023-9-24 16:23:32
            #include <stdio.h> 
            int fun2(int k){
            	int T;
            	T=k*(k+1)/2;
            	return T;}
            
            int fun(int n){
            	int sum=0;
            	for(;n>0;n--)
            	sum+=n*fun2(n+1);
            	return sum;}
            
            int main()
            {
            	int N;
            	scanf("%d",&N);
                int n[N+5];
            	for (int i=0;i<N;i++)
            	scanf("%d",&n[i]);
                for (int i=0;i<N;i++)
                printf("%d %d %d\n",i+1,n[i],fun(n[i]));
            }
            
            • 1

            信息

            ID
            172
            时间
            3000ms
            内存
            128MiB
            难度
            3
            标签
            (无)
            递交数
            304
            已通过
            171
            上传者