题目描述
分析如下代码
for(i=1;i<n;i++)
  for(j=1;j<i;j++)
    for(k=1;k<j;k++)
      printf("\n");
问printf语句共执行了几次？这段代码执行完以后i+j+k值为多少？


输入
由多行组成，每行一个整数n, 1<= n <= 3000
输出
对每一行输入，输出对应的一行，包括空格分开的两个整数，分别代表printf语句的执行次数以及代码执行完以后i+j+k的值, 如果值不确定,输出"RANDOM"取代值的位置


样例输入
6
样例输出
10 15

#include<stdio.h>
int main()
{
    int i,j,k,n,s,m;
    while(scanf("%d",&n)!=EOF)
    {
        if(1<=n&&n<=3000)
        {
            if(n==1)
                printf("0 RANDOM\n");
            if(n==2)
                printf("0 RANDOM\n");
            if(n>=3)
            {
                s=3;
                for(i=3;i<=n;i++)
                    s=s+3;
                    k=0,m=0;
                    for(j=3;j<=n;j++)
                    {
                        k=k+j-3;
                        m=m+k;
                    }
                printf("%d %d\n",m,s);
            }
        }
    }
    return 0;
}
输入n=2347时m就变负数-2145765416了 正确答案是2149201880 直到2956再变正数  怎么回事啊

超出int范围了

#include<stdio.h>
int main()
{
    int i, j,n;
    long long k, s, m;
    while (scanf("%d", &n) != EOF)
    {
        if (1 <= n&&n <= 3000)
        {
            if (n == 1)
                printf("0 RANDOM\n");
            if (n == 2)
                printf("0 RANDOM\n");
            if (n >= 3)
            {
                s = 3;
                for (i = 3; i <= n; i++)
                    s = s + 3;
                k = 0, m = 0;
                for (j = 3; j <= n; j++)
                {
                    k = k + j - 3;
                    m = m + k;
                }
                printf("%lld %lld\n", m, s);
            }
        }
    }
    return 0;
}