【YBTOJ】【Luogu P6218】[USACO06NOV] Round Numbers S

【YBTOJ】【Luogu P6218】[USACO06NOV] Round Numbers S

链接：

洛谷

题目大意：

在 \([l,r]\) 中找到二进制中零数大于等于一数的数的个数。

正文：

数位 DP 板子题。设 \(f_{len,A,B,pos}\) 表示当前 \(len\) 位 \(A\) 个零、\(B\) 个一，碰没碰顶的方案数。

代码：

const int N = 60;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

ll l, r;

int a[N], Len;
ll f[N][N][N][2];
ll DP (int len, int A, int B, bool pos)
{
	if (~f[len][A][B][pos]) return f[len][A][B][pos];
	if (!len) return A >= B;
	int m = pos? a[len]: 1;
	ll ans = 0;
	for (int i = 0; i <= m; i++)
		ans += DP(len - 1, (B > 0) * (A + (i == 0)), B + (i == 1), pos && i == m);
	return f[len][A][B][pos] = ans;
}

ll Solve (ll n)
{
	if (!n) return 1;
	Len = 0;
	memset (a, 0, sizeof a);
	for (ll m = n; m; m /= 2)
		a[++Len] = m % 2;
	ll ans = 0;
	for (int i = 0; i <= a[Len]; i++)
		ans += DP(Len - 1, 0, i == 1, i == a[Len]);
	return ans;
}

int main()
{
	l = Read(), r = Read();
	memset (f, -1, sizeof f);
	printf ("%lld\n", Solve(r) - Solve(l - 1));
	return 0;
}