Given an integer N(0 ≤ N ≤ 1000),your task is to calculate #include #include int a[1000][1000]; int main() { int i,j,n,f=0; memset(a,0,sizeof(a)); a[0][1000]=0; a[1][1000]=1; for(i=2;i=0;j--) a[i][j]=a[i-1][j]*i; for(j=1000;j>=0;j--) if(a[i][j]>=

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Given an integer N(0 ≤ N ≤ 1000),your task is to calculate #include #include int a[1000][1000]; int main() { int i,j,n,f=0; memset(a,0,sizeof(a)); a[0][1000]=0; a[1][1000]=1; for(i=2;i=0;j--) a[i][j]=a[i-1][j]*i; for(j=1000;j>=0;j--) if(a[i][j]>=
Given an integer N(0 ≤ N ≤ 1000),your task is to calculate
#include
#include
int a[1000][1000];
int main()
{
int i,j,n,f=0;
memset(a,0,sizeof(a));
a[0][1000]=0;
a[1][1000]=1;
for(i=2;i=0;j--)
a[i][j]=a[i-1][j]*i;
for(j=1000;j>=0;j--)
if(a[i][j]>=1000)
{
a[i][j-1]+=a[i][j]/1000;
a[i][j]%=1000;
}
}
while(scanf("%d",&n)!=EOF)
{
f=0;
for(j=0;j

Given an integer N(0 ≤ N ≤ 1000),your task is to calculate #include #include int a[1000][1000]; int main() { int i,j,n,f=0; memset(a,0,sizeof(a)); a[0][1000]=0; a[1][1000]=1; for(i=2;i=0;j--) a[i][j]=a[i-1][j]*i; for(j=1000;j>=0;j--) if(a[i][j]>=
你的算法是有问题的!
明显的错误就是数组越界!
我用java给你写了一个,你做下参考:
public class TestPlus {
public TestPlus() {
// TODO Auto-generated constructor stub
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int task = 100;//要计算的目标数
calcN(task);
}
/**
* 计算task的阶乘
* @param task
*/
private static long calcN(int task) {
// TODO Auto-generated method stub
long result = 0;
switch(task){
case 0:
result = 1;break;
default :
result = calc(task);break;
}
System.out.println("==="+result);
return result;
}
private static long calc(int task) {
// TODO Auto-generated method stub
long result = 1;
for(int i=1;i

英语翻译Given any integer 0 If is given that N is an integer.Prove that (a) 3(2N-1)+1 is an even number. Given a positive integer N,you should output the most right digit of N^N. 英语翻译Well,it seems the first problem is too easy.I will let you know how foolish you are later. feng5166 says.The second problem is,given an positive integer N,we define an equation like this:N=a[1]+a[2]+a[3]+...+a[m];a[i]>0,1 求c语言高手做此题Well,it seems the first problem is too easy.I will let you know how foolish you are later. feng5166 says.The second problem is,given an positive integer N,we define an equation like this:N=a[1]+a[2]+a[3]+...+a[m];a[i]>0,1 杭电acm1017题 感觉代码没错,但就是ac不了,我无解了,题目是这样的:Given two integers n and m,count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer.This problem contains multiple 英语翻译Given an integer array A,which contains N integers.Please calculate the sum of f[i].Here,f[i] means the number of integers strictly less than A[i] in A.输入There are several cases.For each case,the first line has an integer N,then follo Given an integer N(0 ≤ N ≤ 1000),your task is to calculate #include #include int a[1000][1000]; int main() { int i,j,n,f=0; memset(a,0,sizeof(a)); a[0][1000]=0; a[1][1000]=1; for(i=2;i=0;j--) a[i][j]=a[i-1][j]*i; for(j=1000;j>=0;j--) if(a[i][j]>= 英语翻译constnum = 5;varn:integer;function r(n :integer) :integer;vari :integer;beginif n a^2+b^2中的^什么意思0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer. Please enter an integer是什么意思 有几道GMAT数学题弄不懂,1、If 1050(是10的50次方) – 74 is written as an integer in base 10 notation,what is the sum of the digits in that integer?2、If x > 0,then 1/[√(2x)+√x] =3、A positive integer n is said to be “prime-sa 离散数学编程题目, 谢谢大牛Given a positive integer n.Choose 3 distinguishing numbers a1,a2,a3 from 1 to n(1 一道数学的ACM题.1514:x + 2y + 3z = nTime Limit:1000MS Memory Limit:65536KTotal Submit:229 Accepted:52 [Submit] [Status] [Discuss] Font Size:Aa Aa Aa DescriptionThis problem is so easy:given an positive integer n,you are to find the number of so integer,integer(),integer(0)的区别求大神解释下 GRE猴哥112数学难题 40题if n is an integer and 99< n的平方 Java write a program to read an integer,n,from the keyboard and output n row of pascal's triangle. VB练习题,,请高手帮我完成 a=(2n+1)!,n is an positive integer