POJ2282：The Counting Problem(数位DP)

Description
Giventwointegersaandb,wewritethenumbersbetweenaandb,inclusive,inalist.Yourtaskistocalculatethenumberofoccurrencesofeachdigit.Forexample,ifa=1024andb=1032,thelistwillbe

102410251026102710281029103010311032

thereareten0'sinthelist,ten1's,seven2's,three3's,andetc.
Input
Theinputconsistsofupto500lines.Eachlinecontainstwonumbersaandbwhere0<a,b<100000000.Theinputisterminatedbyaline`00',whichisnotconsideredaspartoftheinput.
Output
Foreachpairofinput,outputalinecontainingtennumbersseparatedbysinglespaces.Thefirstnumberisthenumberofoccurrencesofthedigit0,thesecondisthenumberofoccurrencesofthedigit1,etc.
SampleInput

110
44497
346542
11991748
14961403
1004503
1714190
1317854
1976494
10011960
00

SampleOutput

1211111111
851851851851909696969593
404040931368240404040
115666215215214205205154105106
1611319201142020191916
107105100101101197200200200200
4131133503503503502502417402412
19651218610487939797142196
3981375398398405499499495488471
2941256296296296296287286286247

题意：求出区间内0~9的个数

思路：dp[i][j]，代表长度为i的数字里面共有几个j

#include<stdio.h>
#include<string.h>
#include<algorithm>
usingnamespacestd;

__int64dp[10][10];

__int64pow(__int64a,__int64b)
{
__int64ans=1,i;
for(i=1;i<=b;i++)
ans*=a;
returnans;
}

voidset()
{
memset(dp,0,sizeof(dp));
for(__int64i=1;i<=9;i++)
{
for(__int64j=0;j<=9;j++)
dp[i][j]=dp[i-1][j]*10+pow(10,i-1);
}
}

__int64solve(__int64n,__int64pos)
{
__int64s[15],len=0,i,tem;
while(n)
{
s[++len]=n%10;
n/=10;
}
s[0]=0;
__int64ans=0,cnt=0;
for(i=len;i>=1;i--)
{
if(pos==0&&i==len)//找出0的情况
{
ans+=dp[i-1][pos]*(s[i]-1);
tem=pow(10,i-1)-1;
ans+=solve(tem,0);
}
else//不为0,则后面的个数要进行s[i]次相加
ans+=dp[i-1][pos]*s[i];
ans+=cnt*pow(10,i-1)*s[i];//计算前面的与要求的数字相同的被运算了几次
if(pos)
{
if(s[i]>pos)
ans+=pow(10,i-1);//s[i]>tmp，则s[i]==tmp时,要将刺猬相加pow(10,i-1)次
if(s[i]==pos)
cnt++;
}
elseif(i!=len)
{
if(s[i]>pos)
ans+=pow(10,i-1);
if(s[i]==pos)
cnt++;

}
}
returnans+cnt;
}

intmain()
{
set();
__int64l,r,i;
while(~scanf("%I64d%I64d",&l,&r),l+r)
{
if(l>r)
swap(l,r);
printf("%I64d",solve(r,0)-solve(l-1,0));
for(i=1;i<=9;i++)
printf("%I64d",solve(r,i)-solve(l-1,i));
printf("\n");
}
return0;
}