• [POJ3252]Round Number(数位dp)


    题目链接:http://poj.org/problem?id=3252

    题意:求范围内数字二进制下0的个数大于等于1的个数的数的个数。

    数位dp,dp(l,zero,one,fz)记录当前第l位时0的个数1的个数和当前位是否是前导零中的部分,dfs转移就行。

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define fr first
     4 #define sc second
     5 #define cl clear
     6 #define BUG puts("here!!!")
     7 #define W(a) while(a--)
     8 #define pb(a) push_back(a)
     9 #define Rint(a) scanf("%d", &a)
    10 #define Rll(a) scanf("%I64d", &a)
    11 #define Rs(a) scanf("%s", a)
    12 #define Cin(a) cin >> a
    13 #define FRead() freopen("in", "r", stdin)
    14 #define FWrite() freopen("out", "w", stdout)
    15 #define Rep(i, len) for(int i = 0; i < (len); i++)
    16 #define For(i, a, len) for(int i = (a); i < (len); i++)
    17 #define Cls(a) memset((a), 0, sizeof(a))
    18 #define Clr(a, x) memset((a), (x), sizeof(a))
    19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
    20 #define lrt rt << 1
    21 #define rrt rt << 1 | 1
    22 #define pi 3.14159265359
    23 #define RT return
    24 #define lowbit(x) x & (-x)
    25 #define onecnt(x) __builtin_popcount(x)
    26 typedef long long LL;
    27 typedef long double LD;
    28 typedef unsigned long long ULL;
    29 typedef pair<int, int> pii;
    30 typedef pair<string, int> psi;
    31 typedef pair<LL, LL> pll;
    32 typedef map<string, int> msi;
    33 typedef vector<int> vi;
    34 typedef vector<LL> vl;
    35 typedef vector<vl> vvl;
    36 typedef vector<bool> vb;
    37 
    38 const int maxn = 34;
    39 int digit[maxn];
    40 LL dp[maxn][maxn][maxn][2];
    41 
    42 int k;
    43 LL l, r;
    44 
    45 LL dfs(int l, int zero, int one, bool fz, bool flag) {
    46   if(l == 0) return zero >= one;
    47   if(!flag && ~dp[l][zero][one][fz]) return dp[l][zero][one][fz];
    48   LL ret = 0;
    49   int pos = flag ? digit[l] : 1;
    50   Rep(i, pos+1) {
    51     if(i == 0 && fz) ret += dfs(l-1, 0, one, fz, flag&&(pos==i));
    52     else ret += dfs(l-1, zero+(i==0), one+(i==1), false, flag&&(pos==i));
    53   }
    54   if(!flag) dp[l][zero][one][fz] = ret;
    55   return ret;
    56 }
    57 
    58 LL f(LL x) {
    59   int pos = 0;
    60   while(x) {
    61     digit[++pos] = x & 1;
    62     x >>= 1;
    63   }
    64   return dfs(pos, 0, 0, true, true);
    65 }
    66 
    67 signed main() {
    68   //FRead();
    69   Clr(dp, -1);
    70   while(cin >> l >> r) {
    71     cout << f(r) - f(l-1) << endl;
    72   }
    73   RT 0;
    74 }
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  • 原文地址:https://www.cnblogs.com/kirai/p/5895205.html
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