Timus Online Judge 1057. Amount of Degrees(数位dp)
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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 详细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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