标签：dp
Lucifer is the only human whi has defeated RA-ONE in a computer game ..
RA-One is after lucifer for revenge and G-One is there to protect him ...
All thi G-One and Ra-one Nonsense has disturbed lucifers life..
He wants to get Rid of Ra-One and kill him . He found that Ra-One can be killed only by throwing Lucifer number of weapons at him.
Lucifer number shares the some properties of Ra-One Numbers numbers and G-One Numbers
Any number is LUCIFER NUMBER if the Difference between Sum of digits at even location and Sum of digits at odd location is prime number .. For eg... for 20314210 is lucifer number
digits at odd location 0,2,1,0
digits at even location 1,4,3,2
diff = (1+4+3+2)-(0+2+1+0)=10-3 = 7 ..... a prime number.
Lucifer has access to a Warehouse which has lots of weapons ..
He wants to know in how many ways can he kill him.
Can you help him?
First line will have a number ‘t‘ denoting the number of test cases.
each of the following t lines will have 2 numbers ‘a‘ , ‘b‘
Print single number per test case, depicting the count of Lucifer numbers in the range a,b inclusive.
Input:
5
200 250
150 200
100 150
50 100
0 50
Output:
2
16
3
18
6NOTE: t will be less than 100
from and to will be between 0 and 10^9 inclusive
/*
链接:http://www.spoj.com/problems/LUCIFER/en/
题意:求一个区间内偶数位置减去奇数位置的差为质数的数的个数
思路:数位dp dp[i][j][k] 第i位，奇数位和为j，偶数位和为k的情况
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define bug printf("hihi\n")
#define eps 1e-8
typedef long long ll;
using namespace std;
#define N 11
int dp[N][90][90];
int bit[N];
int pri[200];
void inint()
{
int i,j;
pri[1]=1;
pri[0]=1;
for(i=2;i<200;i++)
if(!pri[i])
for(j=i*2;j<200;j+=i)
pri[j]=1;
}
int dfs(int pos,int odd,int even,bool bound)
{
if(pos==0)
{
if(even<=odd) return 0;
if(pri[even-odd]) return 0;
return 1;
}
if(!bound&&dp[pos][odd][even]>=0) return dp[pos][odd][even];
int up=bound ? bit[pos]:9;
int ans=0;
for(int i=0;i<=up;i++)
{
int tt;
if(pos&1)
tt=dfs(pos-1,odd+i,even,bound&&i==up);
else
tt=dfs(pos-1,odd,even+i,bound&&i==up);
ans+=tt;
}
if(!bound) dp[pos][odd][even]=ans;
return ans;
}
int solve(int x)
{
int i,j,len=0;
while(x)
{
bit[++len]=x%10;
x/=10;
}
return dfs(len,0,0,true);
}
int main()
{
int i,j,t;
memset(dp,-1,sizeof(dp));
inint();
scanf("%d",&t);
int x,y;
while(t--)
{
scanf("%d%d",&x,&y);
printf("%d\n",solve(y)-solve(x-1));
}
return 0;
}
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标签：dp
原文地址：http://blog.csdn.net/u014737310/article/details/47040067
