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1057. Amount of Degrees
Time limit: 1.0 second
Memory limit: 64 MB Create a code to determine the amount of integers, lying in the set [X;Y] and being a sum of exactlyK different integer degrees of B.
Example. Let X=15, Y=20, K=2, B=2. By this example 3 numbers are the sum of exactly two integer degrees of number 2:
17 = 24+20,
18 = 24+21, 20 = 24+22. InputThe first line of input contains integers X and Y, separated with a space (1 ≤ X ≤ Y ≤ 231?1).
The next two lines contain integers K and B (1 ≤ K ≤ 20; 2 ≤ B ≤ 10).
OutputOutput should contain a single integer — the amount of integers, lying between X and Y, being a sum of exactly K different integer degrees of B.
Sample
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/*
题意: 求一个区间的 degree进制的1的个数为k的数的个数
思路:数位dp,一定要注意是1个个数为k dp[i][j][k] 代表到达了i位的j进制还差k个1
具体注意的地方写在了代码中
*/
#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 35
int dp[33][15][33];
int degree,k;
int bit[N];
int dfs(int pos,int degree,int t,bool bound)
{
if(t<0) return 0;
if(pos==0) return t ? 0:1;
if(!bound&&dp[pos][degree][t]>=0) return dp[pos][degree][t];
int up=bound ? min(bit[pos],1):1;
int ans=0;
for(int i=0;i<=up;i++)
ans+=dfs(pos-1,degree,t-i,bound&&i==bit[pos]); //必须是bit[pos]，不能是uo
if(!bound) dp[pos][degree][t]=ans;
return ans;
}
int solve(int x)
{
int i,j;
int len=0;
while(x)
{
bit[++len]=x%degree;
x/=degree;
}
return dfs(len,degree,k,true);
}
int main()
{
int i,j,le,ri;
memset(dp,-1,sizeof(dp));
while(~scanf("%d%d",&le,&ri))
{
scanf("%d%d",&k,°ree);
printf("%d\n",solve(ri)-solve(le-1));
}
return 0;
}
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Timus Online Judge 1057. Amount of Degrees（数位dp）
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原文地址：http://blog.csdn.net/u014737310/article/details/47053413
