求一正整数等差数列的前六项的和，该数列前四项之和是二十六，四项之积是八百八

设首项为a1，公差为d,由题意有4×（a1）+6d=26，
(a1)×（a1+d）×（a1+2d）×（a1+3d)=880
又因为该数列是正整数等差数列
所以解得a1=2,d=3
所以该数列前六项的和S6=6(a1)+15d=57

#include<stdio.h>
main()
{
int a[6],i,step;
for(a[0]=1;;a[0]++)
{
for(step=1;step<4;step++)
{
for(i=1;i<6;i++)
{
a[i]=a[i-1]+step;
}
if(a[0]+a[1]+a[2]+a[3]==26&&a[0]*a[1]*a[2]*a[3]==880)
{
printf("%d\n",a[0]*a[0]+a[1]*a[1]+a[2]*a[2]+a[3]*a[3]+a[4]*a[4]+a[5]*a[5]);
printf("%d,%d",a[0],step);
return(0);
}
}
}
}

a2 + a3 = a1 + a4 = 26/2 = 13 = 2a2 + d,a1a2a3a4 = a2·(a2+d)(a2-d)(a2 + 2d) = 880,联立解得：a2 = 5,d = 3,a1=5-3 = 2,a5 = 5+3*3 = 14,a6 = a5+3 = 17,S6 = 26+14+17 = 57

首项为2，公差为3，正整数等差数列的前六项的和57