标签：
题意：给一个数q，
q=1时求给定区间，给定进制，各数位和等于m的数字的个数
q=2时求给定区间，给定进制，各数位和等于m的数字中的第k大的数字
分析：dp[i][sum][j],表示长度为i当前数位和是sum,进制是j的个数，q=2时用二分求出k大数
题意给的区间[x,y]，x不一定小于y，给定区间没k大数，则输出 Could not find the Number!
#include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <string> #include <cctype> #include <complex> #include <cassert> #include <utility> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef pair<int,int> PII; typedef long long ll; #define lson l,m,rt<<1 #define pi acos(-1.0) #define rson m+1,r,rt<<11 #define All 1,N,1 #define read freopen("in.txt", "r", stdin) const ll INFll = 0x3f3f3f3f3f3f3f3fLL; const int INF= 0x7ffffff; const int mod = 1000000007; ll dp[50][500][65],x,y,k; int b,m,bit[50]; ll dfs(int i,int s,int j,int e){ if(i==0)return (s==m); if(!e&&dp[i][s][j]!=-1)return dp[i][s][j]; int u=e?bit[i]:b-1; ll num=0; for(int v=0;v<=u;++v){ num+=dfs(i-1,s+v,j,e&&(v==u)); } return e?num:dp[i][s][j]=num; } ll solve1(ll a){ int len=0; if(a<0)return 0; while(a){ bit[++len]=a%b; a/=b; } return dfs(len,0,b,1); } ll solve2(){ ll l=x,r=y; ll num=solve1(x-1); if(solve1(y)-num<k)return -1; while(l<=r){ ll mid=(l+r)>>1; if(solve1(mid)-num<k)l=mid+1; else r=mid-1; } return l; } int main() { int q,cas=0; while(~scanf("%d",&q)){ memset(dp,-1,sizeof(dp)); printf("Case %d:\n",++cas); if(q==1){ scanf("%I64d%I64d%d%d",&x,&y,&b,&m); if(x>y)swap(x,y); printf("%I64d\n",solve1(y)-solve1(x-1)); } else{ scanf("%I64d%I64d%d%d%I64d",&x,&y,&b,&m,&k); if(x>y)swap(x,y); ll tmp=solve2(); if(tmp!=-1) printf("%I64d\n",tmp); else printf("Could not find the Number!\n"); } } return 0; }
标签：
原文地址：http://www.cnblogs.com/zsf123/p/4680844.html
