P4399P4999
It has three times the experience is a treasure topics
We can find out the 1 to n, 1 to 9 appeared several times, provided f [i] is the number of number i appears, the \ (ans = \ sum {f [i] \ cdot i} \)
Then is the number of bits dp doing things
We can count the number of times the sum of the current requirement occurs when the number of goal dp, return sum when you can reach the border. Note considerations leading zeros
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int inf=214748364;
const ll mod=1000000007;//注意是1e9+7
inline ll read()
{
char ch=getchar();
ll x=0;bool f=0;
while(ch<'0'||ch>'9')
{
if(ch=='-')f=1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=(x<<1)+(x<<3)+(ch^48);
ch=getchar();
}
return f?-x:x;
}
ll l,r;
ll f[10],g[20][250],li[20];
int t;
void make(ll k)
{
t=0;
while(k)
{
li[++t]=k%10;
k/=10;
}
}
ll dfs(int now,int sum,int goal,bool lim,bool al)//al记录是否前面全是0
{
if(!now)return sum;
if(!al&&!lim&&g[now][sum]!=-1) return g[now][sum];
int up=lim?li[now]:9;
ll rtn=0;
for(int i=0;i<=up;i++)
{
rtn+=dfs(now-1,sum+((i||(!al))&&(i==goal)),goal,lim&&(i==up),al&&(!i));//这里sum的计算方式是在考虑goal是0的时候，排除前导零的影响（在另外那两倍经验那里也适用）
rtn=(rtn+mod)%mod;
}
if(!lim&&!al) g[now][sum]=rtn;
return rtn;
}
int main()
{
int fk=read();
while(fk--)
{
ll ans=0;
l=read();
r=read();
make(r);
for(int i=1;i<=9;i++)
memset(g,-1,sizeof(g)),f[i]=dfs(t,0,i,1,1);
if(l>1)
{
make(l-1);
for(int i=1;i<=9;i++)
memset(g,-1,sizeof(g)),f[i]-=dfs(t,0,i,1,1);
}
for(int i=1;i<=9;i++)
ans=(ans+((f[i]%mod)*i+mod)%mod+mod)%mod;
printf("%lld\n",ans);
}
}In addition to twice the experience is P2602 and P1239
Manual funny