Question

编程求s=1*2+2*3+3*4+……+n*(n+1），当s>200000时n的最小值

Answer 1

等一下！
#include<stdio.h>
main()
{
int n=0,t=0,s=0;
for(n=1;;n++)
{ t=n*(n+1);
s=s+t;
if(s>200000) break;
}
printf("%d",n);
}

Answer 2

解：s=0,n=1
if s>200 000
print n
else n=n+1
m=n*(n+1)
s=s+m
endif
end

Answer 3

#include <stdio.h>

int main()
{
int n = 0;
int s = 0;

while(s<200000)
{
n++;
s+=n * (n+1);
}
printf("%d\n",n);
}

Answer 4

double s=0;

while(s<200000)
{
s+=n*(n+1);
n++;
}
printf("%d",n-1);

Answer 5

#include <stdio.h>

int main()
{
int n = 1;
int s = 0;

do
{
s = s + n * (n+1);
n++;
}
while (s < 200000);

printf("%d\n",n);
}

结果是：85