A number is called bi-peak if it is a concatenation of two peak numbers.
The score of a number is the sum of all digits. Love8909 is crazy about bi-peak numbers. Please help him to calculate the MAXIMUM score of the Bi-peak Number in the closed interval [A, B].
Each case consists of two integers “A B” (without quotes) (0 <= A <= B < 2^64) in a single line.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 99999999
#define MOD 1000000007
int wei1[100],wei2[100];
int dp[100][10][8];
int dfs(int pos,int pre,int state,int flag1,int flag2)
{
int i,j;
if(pos==0){
if(state==6)return 0;
return -inf;
}
if(!flag1 && !flag2 && dp[pos][pre][state]!=-1){
return dp[pos][pre][state];
}
int min1=flag1?wei1[pos]:0;
int max1=flag2?wei2[pos]:9;
int ans=-inf;
for(i=min1;i<=max1;i++){
if(state==0){
if(i>0){
ans=max(ans,i+dfs(pos-1,i,1,flag1&&i==min1,flag2&&i==max1) );
}
else if(i==0)ans=max(ans,0+dfs(pos-1,i,0,flag1&&i==min1,flag2&&i==max1));
}
else if(state==1){
if(i>pre){
ans=max(ans,i+dfs(pos-1,i,2,flag1&&i==min1,flag2&&i==max1));
}
}
else if(state==2){
if(i>pre){
ans=max(ans,i+dfs(pos-1,i,2,flag1&&i==min1,flag2&&i==max1));
}
if(i<pre){
ans=max(ans,i+dfs(pos-1,i,3,flag1&&i==min1,flag2&&i==max1 ) );
}
}
else if(state==3){
if(i<pre){
ans=max(ans,i+dfs(pos-1,i,3,flag1&&i==min1,flag2&&i==max1 ) );
}
if(i>0){
ans=max(ans,i+dfs(pos-1,i,4,flag1&&i==min1,flag2&&i==max1 ) );
}
}
else if(state==4){
if(i>pre){
ans=max(ans,i+dfs(pos-1,i,5,flag1&&i==min1,flag2&&i==max1));
}
}
else if(state==5){
if(i>pre){
ans=max(ans,i+dfs(pos-1,i,5,flag1&&i==min1,flag2&&i==max1));
}
if(i<pre){
ans=max(ans,i+dfs(pos-1,i,6,flag1&&i==min1,flag2&&i==max1 ) );
}
}
else if(state==6){
if(i<pre){
ans=max(ans,i+dfs(pos-1,i,6,flag1&&i==min1,flag2&&i==max1) );
}
}
}
if(!flag1 && !flag2){
dp[pos][pre][state]=ans;
}
return ans;
}
int main()
{
ull m,n;
int T,i,j,cas=0;
memset(dp,-1,sizeof(dp));
scanf("%d",&T);
while(T--)
{
cin>>m>>n;
int len=0;
while(n){
len++;
wei2[len]=n%10;
n/=10;
wei1[len]=m%10;
m/=10;
}
int ans=dfs(len,0,0,1,1);
if(ans<0)ans=0;
printf("Case %d: %d\n",++cas,ans);
}
return 0;
}