标签：数位dp
In the War between good and evil . Ra-One is on the evil side and G-One on the good side.
Ra-One is fond of destroying cities and its G-one‘s duty to protect them..
Ra-One loves to destroy cities whose Zip Code has special properties. He says he loves to destroy cities which have Ra-One numbers as their ZIp Code.
Any number is Ra-one if the Difference between Sum of digits at even location and Sum of digits at odd location is One (1).. For eg... for 234563 is Ra-One number
digits at odd location are 3,5,3 (unit place is location 1 )
digits at even location are 2,4,6
Diff = (2+4+6)-(3+5+3)=12-11 = 1.
And 123456 is not Ra-One number
diff = (5+3+1) - (2+4+6) = -4
G-One knows this about Ra-one and wants to deploy his Army members in those cities. 1 army member will be deployed in each city.
G-one knows the range of ZIP-Codes where Ra-One might attack & needs your help to find out how many army members he needs.
Can you help Him ?
first line will have only one integer ‘t‘ number of Zip-Code ranges. it is followed by t lines
each line from 2nd line cotains 2 integer ‘from‘ and ‘to‘. these indicate the range of Zip codes where Ra-one might attack .(from and to are included in the range)
A single number for each test case telling how many army members G-One needs to deploy.
each number should be on separate lines
Input: 2
1 10
10 100
Output:
1
9
explanation:
for 1st test case the only number is 10
for 2nd test case numbers are 10,21,32,43,54,65,76,87,98
NOTE: t will be less than 100
from and to will be between 0 and 10^8 inclusive
题意:就是在区间中偶数位和减去奇数位和等于1的数有多少个
思路：直接数位dp记忆化
#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 eps 1e-8
//typedef __int64 ll;
#define fre(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi\n")
using namespace std;
#define mod 1000000007
#define INF 0x3f3f3f3f
#define N 500
int dp[10][N];
int bit[N];
int dfs(int pos,int va,bool bound)
{
if(pos==0) return va==1;
if(!bound&&dp[pos][va]!=-1) return dp[pos][va];
int up=bound ? bit[pos]:9;
int i,t;
int ans=0;
fre(i,0,up+1)
{
if(pos&1) t=-1;
else t=1;
ans+=dfs(pos-1,va+t*i,bound&&i==up);
}
if(!bound) dp[pos][va]=ans;
return ans;
}
int solve(int x)
{
int len=0;
while(x)
{
bit[++len]=x%10;
x/=10;
}
return dfs(len,0,true);
}
int main()
{
mem(dp,-1);
int le,ri;
int t;
sf(t);
while(t--)
{
sff(le,ri);
pf("%d\n",solve(ri)-solve(le-1));
}
pf("\n");
}
spoj RAONE - Ra-One Numbers （数位dp）
标签：数位dp
原文地址：http://blog.csdn.net/u014737310/article/details/45126517
