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POJ 3252 Round Numbers (数位DP)

题意:求区间内一个数二进制位1的数量大于等于0的数的个数。

析:dp[i][j][k] 表示前 i 位,长度为 j 的,1的数量是 k。注意前导0.

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <sstream>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 3e3 + 5;const int mod = 1e9 + 7;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}int dp[35][35][35];int a[35];int dfs(int pos, int num, int val, bool is, bool ok){    if(pos < 0)  return num - val >= val;    int &ans = dp[pos][num][val];    if(!ok && ans >= 0)  return ans;    int res = 0, n = ok ? a[pos] : 1;    for(int i = 0; i <= n; ++i)        res += dfs(pos-1, is&&!i?num-1:num, i?val+1:val, is && !i, ok && i == n);    return ok ? res : ans = res;}int solve(int n){    int len = 0;    for(int i = 0; i < 31; ++i)        if((1<<i)&n){  a[i] = 1; len = i+1; }        else a[i] = 0;    return dfs(len-1, len, 0, true, true);}int main(){    memset(dp, -1, sizeof dp);    while(scanf("%d %d", &m, &n) == 2){        printf("%d\n", solve(n) - solve(m-1));    }    return 0;}

 

POJ 3252 Round Numbers (数位DP)