sorry，I know why...It's the high-percision I designed is wrong...

I store each 3 digit of the orginal number in a int,but when I output the number,I forgot to insert 0 before the int which digit is less than 3.

sorry for my english...

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>

char* add(char *p1, char *p2)
{
    char *p;
    char temp=0, end1=0, end2=0;
    int len,i;
    len=strlen(p1)>strlen(p2)?strlen(p1)+2:strlen(p2)+2;
    p=(char *)malloc(len*sizeof(char));
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(!p2[i]) end2=1;
        if(end1&&end2&&!temp) break;
        if(!end1) temp+=(p1[i]-'0');
        if(!end2) temp+=(p2[i]-'0');
        p[i]=temp%10+'0';
        temp/=10;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

char* sub(char *p1, char *p2)
{
    char *p;
    char temp=0, end1=0, end2=0;
    int len,i;
    len=strlen(p1)>strlen(p2)?strlen(p1)+2:strlen(p2)+2;
    p=(char *)malloc(len*sizeof(char));
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(!p2[i]) end2=1;
        if(end1&&end2&&!temp) break;
        if(!end1) temp+=(p1[i]-'0');
        if(!end2) temp-=(p2[i]-'0');
        p[i]=(temp+100)%10+'0';
        if(temp<0)
        {
            temp/=10;
            temp--;
        }
        else temp=0;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

char* mul(char *p1, int k)
{
    char *p;
    char temp=0, end1=0;
    int len,i;
    len=strlen(p1)+2;
    p=(char *)malloc(len*sizeof(char));
    if(k==10)
    {
        p[0]='0';
        p[1]=0;
        strcat(p,p1);
        return p;
    }
    for(i=0; i<len-1; ++i)
    {
        if(!p1[i]) end1=1;
        if(end1&&!temp) break;
        if(!end1) temp+=(p1[i]-'0')*k;
        p[i]=temp%10+'0';
        temp/=10;
    }
    p[i]=0;
    while(p[--i]=='0'&& &p[i]!=p) p[i]=0;
    return p;
}

void printb(char *p)
{
    int i;
    for(i=strlen(p)-1;i>=0; i--) printf("%c", p[i]);
}

int main()
{
    int N, K;
    int i;
    char* result, *t1, *t2;
    char* Nulls[2000];

    printf("Enter N and K: ");
    scanf("%d%d",&N,&K);

    Nulls[0]="0";
    result=(char *)malloc(2*sizeof(char));
    result[0]=K-1+'0';
    result[1]=0;

    try
    {

    for(i=1; i<N;++i)
    {
        Nulls[i]=sub(result,Nulls[i-1]);
        t1=mul(Nulls[i],K);
        t2=mul(Nulls[i-1],(K-1));
        result=add(t1,t2);
        free(t1);
        free(t2);
    }

    }
    catch(...)
    {
        printf("Error");
        while(!kbhit());
        return 0;
    }

    printf("Result = ");
    printb(result);
    free(result);
    while(!kbhit());
    return 0;
}

kazaxam respect!

Edited by author 08.02.2006 16:43

Edited by author 08.02.2006 16:43

longint is not enough

竟让可以打中文？？！！》》》》》》》

Use base = 10
compare small answers with 1009

Чёрт его знает

Помоему он подсовывает во 2 тесте N=170 и K=10 Если умножать пллучается 10 с 170 нулями. Никакой тип не выдержит. Надо разбивать число на  группы. Только какие?

Edited by author 11.06.2008 17:47

174568663359808123517619015636369512622213
064344534689513998708986681415716975079951
752754958954321404709642918328732491072229
658639887075761909680804620151071937764270
946419913401

Edited by author 29.10.2008 11:30

in my programm I added this:
if(k==1)
{cout<<0;return 0;}
if(n==1)
{cout<<k-1;return 0;}

according to problem statement,n>=2,k>=2,so this cases are impossible,but here I got WA1,when without this cases I got WA2!!

Thanks you !!!

Try to solve it with long arithmetics... Delete if's and then in a unique repeat function use long arithmetics and that's all.don't increase t just bind t:=j; and you will get AC...
also the sizes of arrays are 2000 and you should print the 2nd array i mean f2 not f3. good luck!

P.S. if you have questions i'll be glad to help you...

You are wrong.

"k-based number" means "K进制数"
"successful zeroes" means "连续的零"

Because you are very ugly!
                ~~~~

Edited by author 21.07.2007 16:48

You way are currect but you MUST use HI-precision

You should use long arithmetic!!!

var
  a,b:array[1..1000]of longint;
  i,n,k,u,v,j:longint;
begin
  for i:=1 to 100 do
  begin
    a[i]:=0;
    b[i]:=0;
  end;
  readln(n);
  readln(k);
  a[1]:=k-1;
  b[1]:=k*(k-1);
  u:=1;
  v:=1;
  for i:=3 to n do
  if (i mod 2=1) then
  for j:=1 to v do
  begin
  if 9*(a[j]+b[j])>=1000000000
  then begin
  if j=v then u:=u+1;
  a[j+1]:=a[j+1]+1;
  end;
  a[j]:=9*(a[j]+b[j]) mod 1000000000
  end
  else for j:=1 to u do
  begin
  if 9*(a[j]+b[j])>=1000000000
  then begin
  if j=u then v:=v+1;
  a[j+1]:=a[j+1]+1;
  end;
  a[j]:=9*(a[j]+b[j]) mod 1000000000
  end;
  if n mod 2=1 then
  for i:=u downto 1 do
    write(a[i])
  else for i:=v downto 1 do
    write(b[i]);
end.