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题目链接:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1766 Accepted Submission(s): 669
#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; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5954815.html