hdu-5491 The Next(贪心)
题目链接:
The Next
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1766 Accepted Submission(s): 669
Problem Description
Let L denote the number of 1s in integer D’s binary representation. Given two integers S1 and S2, we call D a WYH number if S1≤L≤S2.
With a given D, we would like to find the next WYH number Y, which is JUST larger than D. In other words, Y is the smallest WYH number among the numbers larger than D. Please write a program to solve this problem.
With a given D, we would like to find the next WYH number Y, which is JUST larger than D. In other words, Y is the smallest WYH number among the numbers larger than D. Please write a program to solve this problem.
Input
The first line of input contains a number T indicating the number of test cases (T≤300000).
Each test case consists of three integers D, S1, and S2, as described above. It is guaranteed that 0≤D<231 and D is a WYH number.
Each test case consists of three integers D, S1, and S2, as described above. It is guaranteed that 0≤D<231 and D is a WYH number.
Output
For each test case, output a single line consisting of “Case #X: Y”. X is the test case number starting from 1. Y is the next WYH number.
Sample Input
3
11 2 4
22 3 3
15 2 5
Sample Output
Case #1: 12
Case #2: 25
Case #3: 17
题意:
给一个D,现在让求最小的ans>D,且ans的二进制中1的个数是[s1,s2];
思路:
套路题,从高位到低位按位贪心,对于当前位,如果为1,那么这一位就一定要取1,然后d,s1,s2都变小了;
如果当前位为0,那么就要看取最多个数的1能得到的数如果大于当前的数,那么还可以取0;
否则就要取1了;就这样贪心,具体的可以看代码;
AC代码:
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int maxn=1e5+10; int l,r; LL d,ans,dp[40]; void dfs(LL cur,int hi,int lo,int pos) { if(hi<=0||pos<0)return ; if(cur>=dp[pos]){ans+=dp[pos],dfs(cur-dp[pos],hi-1,lo-1,pos-1);} else { if(lo==pos+1) { ans+=dp[lo]-1; return ; } else { LL num=0; for(int i=pos-1;i>=max(0,pos-hi);i--)num+=dp[i]; if(num>cur)dfs(cur,hi,lo,pos-1); else { ans+=dp[pos]; lo--; if(lo>0)ans+=dp[lo]-1; return ; } } } } int main() { int t,Case=0; scanf("%d",&t); dp[0]=1; for(int i=1;i<=33;i++)dp[i]=dp[i-1]*2; while(t--) { scanf("%lld%d%d",&d,&l,&r); ans=0; dfs(d,r,l,32); if(d==0) { if(l==0)ans=1; else ans=dp[l]-1; } printf("Case #%d: %lld\n",++Case,ans); } return 0; }