Function f(n) is recursively defined as:

•    f(n) = f(n-1) + f(n-3), n > 3
•    f(n) = n, n <= 3

Write a program to calculate f(n) modulo m.
n <= 10000, 2 <= m <= 10000.

Input
There are multiple test cases. Each test case consists of two integers: n and m. n = 0 and m = 0 denotes the end of input, and you should not process this case.

Output
For each test case, print f(n) modulo m in a single line.

Sample Input
1 2
10000 999
0 0

Sample Output
1
433

怎么做
我是菜鸟啊
不会啊

#include <stdio.h>
#define N 100000
int main()
{
    int a[N];
    int m,n;
    while(scanf("%d%d",&n,&m)&&(m||n))
    {
        if((n<1)||(n>10000)||(m>10000)||(m<2))
        {
            fflush(stdin);
            printf("error,please input again!\n");
            continue;
        }
        else if(n<=3)
        {
            printf("%d\n",n%m);
        }
        else
        {
            a[1]=1;
            a[2]=2;
            a[3]=3;
            int i;
            for( i=4;i<=n;i++)
                a[i]=a[i-1]+a[i-3];
            printf("%d\n",a[i-1]%m);
        }
    }
}

结果运算不对啊

f（10000） 以超出了整形的范围！~

输入10000 999
结果不是433

还没懂啊！f（10000） 太大了 超出整形的范围！你可以改成double 型自己写啊

谢谢啊