已知一个数列的前三项为0,0,1,以后各项都是其相邻前三项的的和,求该数列前30项之和.

不难啊，数列你可以用数组解决啊a[30]或a[29],把前三项赋值，for(i=1;i<=30;i+c>)wile(i＞=4){a[i]=a[i－1]+[i－2]+a[i－3];
sum=sum+a[i]}
思想是这样。具体代码自己完善吧

#include "math.h"
main()
{ int a[30],sum,i=3;
a[0]=0;
a[1]=0;
a[2]=1;
sum= a[0]+ a[1]+a[2];
while(i<30)
{ a[i]= a[i-1]+a[i-2]+a[i-3];
sum=sum+a[i];
i++;
}
printf("%d",sum);
getch();
}
结果为7840

0,0,1,1,2,4,7,13,24,44,........
int
main()
{
int
sum
=
0;
int
a[10]
=
{0};
a[0]
=
0;
a[1]
=
0;
a[2]
=
1;
int
i;
for
(i
=
3;
i
<
10;
i
++)
{
a[i]
=
a[i-1]
+
a[i-2]
+
a[i-3];
}
for
(i
=
0;
i
<
10;
i
++)
{
sum
+=
a[i];
printf("
%d
",
a[i]);
}
printf("sum
=
%d\n",
sum);
return
0;
}

#include
int
main()
{
long
a[30],i;
a[0]=a[1]=0;a[2]=1;
printf("0\n0\n1\n");
for(i=3;i<30;i++)
{

a[i]=a[i-1]+a[i-2]+a[i-3];

printf("%ld\n",a[i]);
}
return
0;
}

int a[30]={0,0,1};
int i=3;
int sum=1;

while(i<30)
{
a[i]=a[i-1]+a[i-2]+a[i-3];
sum+=a[i];
i++;
}