程序改错3

改错题目（启动VF，按照题目的要求，改正错误，得出正确的结果）
1、"设某四位数的各位数字的立方和等于100，问满足这种条件的四位数且是素数的数共有多少个？‘4’
set talk off
clea
n=0
for x=1000 to 9999
f=1
for i=2 to int(sqrt(x))
if mod(i,x)=0
f=0
exit
endif
endfor
if f=0
a=int(x/1000)
b=int((x-a*1000)/100)
c=int(mod(x,100)/10)
d=mod(x,10)
if a*a*a+b*b*b+c*c*c+d*d*d=100
n=n+1
endif
endif
endfor
? n
set talk on
return
2、"把一张一元钞票,换成一分、二分和五分硬币,每种至少11枚,问有多少种方案?‘13’
set talk off
clea
n=0
for yi=11 to 100
for er=11 to 20
for wu=11 to 20
if yi+er+wu=100
n=n+1
endif
endfor
endfor
endfor
? n
set talk on
return
3、"找满足以下条件: X^2+Y^2+Z^2=41^2 且X+Y+Z之值最大的三个正整数X,Y,Z, 求X+Y+Z之值.‘71’
set talk off
clea
max=0
for x=0 to 41
for y=0 to 41
for z=0 to 41
if x*2+y*2+z*2=41*41
if max>x+y+z
max=x+y+z
endif
endif
endfor
endfor
endfor
? max
set talk on
return
4、"已知24有8个正整数因子(即:1,2,3,4,6,8,12,24)，而24正好被其因子个数8整除。

求[200,500]之间能被其因子数目整除的数的个数。

‘28’
set talk off
clea
nu=0
for n=200 to 500
s=0
for i=1 to n
if mod(i,n)=0
s=s
endif
endfor
if n>s
nu=nu+1
endif
endfor
? nu
set talk on
return
6、求100以内最小的自然数n，使得从1开始的连续n个自然数的平方之和大于5500。

‘25’
set talk off
clear
s=0
for n=1 to 100
s=s+n*n
if s<=5500
exit
endif
endfor
? s
set talk on
return
7、求出[100,1000]内所有能同时被7和9整除的整数之和。

","7479"
set talk off
clear
s=1
for x=100 to 1000
if mod(x,7)=1 or mod(x,9)=0
s=s+1
endif
endfor
? s
set talk on
return
8、将大于1000且能被3和5中至少一个数整除的所有整数按从小到大顺序排列后，求前面20个数之和。

‘20465’
set talk off
clear
k=0
s=0
x=1000
do while k<=20 &&
x=x+1
if mod(x,3)=0 and mod(x,5)=0 &&
s=s+x
k=k+2 &&
endif
enddo
? s
set talk on
return
==================================================================== ====================================================================
9、求100以内最大的自然数n，使得从1开始的连续n个自然数的倒数之和小于3.5。

‘18’
set talk off
clear
s=0
for n=1 to 100
s=(s+1)/n
if s>=3.5
loop
endif
endfor
? n-1
set talk on
return
"求[101,299]内所有能被2整除但不能同时被3和5整除的整数之和。

《一级》
set talk off
clear
s=0
for x=101 to 299
if mod(x,2)=0 and not mod(x,3)=0 and mod(x,5)=0
x=s+x
endif
endfor
? s
set talk on
return
","18630
求50以内最大的自然数n，使得从102开始的连续n个偶数之和小于3000。

《一级》
set talk off
clear
s=0
for n=1 to 50
a=100+2*n
s=s*a
if s>=3000
loop
endif
endfor
? n-1
set talk on
return
","23
求出[1234,6789]内中间两位数字之和等于首尾两位数字之积的数的个数。

《一级》
set talk off
clear
n=0
for x=1234 to 6789
a=int(x/1000)
b=int(x-a*1000)/100
c=int(x-a*1000-b*100)/10
d=x-a*1000-b*100-c*10
if b+c=a*d
n=n+1
endif
endfor
? n
set talk on
return
","
175"
已知数列{f(n)}：f(1)=1，f(2)=4，当n=3,4,5,…时，f(n)=3*f(n-1)-f(n-2)。

试求出100以内最大的自然数n使得此数列的第n项小于5364321。

《一级》
set talk off
clear
dimension f(100)
f(1)=1
f(2)=4
for n=3 to 100
f(n)=3*f(n-1)-f(n-2)
if f(n)>5364321
loop
endif
endfor
? n
set talk on
return
","
16
"。