洛谷 P6218

题目链接: P6218 [USACO06NOV] Round Numbers S

题目大意

详见题目

solution

有一个显而易见的结论

发现 (ans_{l, r} = ans_{1. r} - ans_{1, l - 1})

那只需要处理 (1 - r) 的即可

设 (f_{i, j, k}) 表示长度为 (i) 最高位为 (j) 的且 (1) 的个数为 (k) 的数的个数

1. (j=1, f_{i, j, k} = sumlimits_{p = 0}^{p leqslant 1}sumlimits_{k = 0}^{k leqslant i} f_{i - 1, p, k - 1})
2. (j=0, f_{i, j, k} = sumlimits_{p = 0}^{p leqslant 1}sumlimits_{k = 0}^{k leqslant i} f_{i - 1, p, k})

对于(ans_{1, r}) 我们可以采用以下策略 :

设 (len) 为 (r) 的位数, (a_{len}) 为 (r) 的每一位

1. 对于首位为 (1) 且前面有 (cnt1) 个 (1) 的 (f) , (res += f_{i, 0, j}, i in [1, len), j in [0, len / 2 - cnt1])
2. 对于首位为 (0) 的 (f) , (res += f_{i, 1, j}, i in [1, len), j in [0, i / 2])

那么答案就是 (ans_{1, r} - ans_{1, l - 1})

Code:

/**
*    Author: Aliemo
*    Data: 
*    Problem: 
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define ll long long
#define rr register

#define inf 1e9
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int l, r, len;

int f[50][2][50], a[50];

inline void init() {
	f[1][1][1] = 1, f[1][0][0] = 1;
	for (rr int i = 2; i < 35; i++) 
		for (rr int j  = 0; j <= 1; j++) 
			for (rr int k = 0; k <= i; k++)
				for (rr int p = 0; p <= 1; p++) 
					if (!j) f[i][j][k] += f[i - 1][p][k];
					else if (k) f[i][j][k] += f[i - 1][p][k - 1];
}

inline int solve(int x) {
	memset(a, 0, sizeof a);
	len = 0;
	while (x) {
		a[++len] = x % 2;
		x /= 2;
	}
	int ans = 0, cnt1 = 1, cnt0 = 0;
	for (rr int i = len - 1; i >= 1; i--) {
		int x = a[i];
		if (x) {
			for (rr int j = 0; j <= len / 2 - cnt1; j++) {
				ans += f[i][0][j];
				// cout << ans << "
";
			}
		}
		cnt1 += x;
		cnt0 += (x == 0);
		if (cnt0 >= cnt1 && i == 1) ans++;
	}
	for (rr int i = 1; i < len; i++)
		for (rr int j = 0; j <= i / 2; j++)
			ans += f[i][1][j];
	return ans;
}

signed main() {
	init();
	l = read(), r = read();
	cout << solve(r) - solve(l - 1);
}