[洛谷P2602][ZJOI2010]数字计数

最新推荐文章于 2025-09-07 19:33:12 发布

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最新推荐文章于 2025-09-07 19:33:12 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/Memory-of-winter/p/9888551.html

本文介绍了一种使用数位动态规划(DP)解决区间内数字计数问题的方法。通过递归函数和节点结构体，实现了高效计算指定区间内各数字0至9出现的次数，特别适用于大范围数值的快速查询。

题目大意：求区间$[l,r]$中数字$0\sim9$出现个数

题解：数位$DP$

卡点：无

C++ Code：

#include <cstdio>
#include <iostream>
struct node {
	long long s[10], sum;
	inline node() {for (register int i = 0; i < 10; i++) s[i] = 0; sum = 0;}
	inline node(int x) {for (register int i = 0; i < 10; i++) s[i] = 0; sum = x;}
	inline friend node operator + (const node &lhs, const node &rhs) {
		node res;
		res.sum = lhs.sum + rhs.sum;
		for (register int i = 0; i < 10; i++) res.s[i] = lhs.s[i] + rhs.s[i];
		return res;
	}
	inline friend node operator - (const node &lhs, const node &rhs) {
		node res;
		res.sum = lhs.sum - rhs.sum;
		for (register int i = 0; i < 10; i++) res.s[i] = lhs.s[i] - rhs.s[i];
		return res;
	}
	inline friend std::ostream & operator << (std::ostream &Fout, const node __node) {
		for (register int i = 0; i < 10; i++) {
			Fout << __node.s[i];
			Fout << (i == 9 ? '\n' : ' ');
		}
		return Fout;
	}
};

int num[20], tot;
node f[20];
bool vis[20];
node calc(int x, int lim, int lead) {
	if (!x) return node(1);
	if (!lim && lead && vis[x]) return f[x];
	node F;
	for (int i = lim ? num[x] : 9, op = 1; ~i; i--, op = 0) {
		node tmp = calc(x - 1, lim && op, lead || i);
		F = F + tmp;
		if (i || lead) F.s[i] += tmp.sum;
	}
	if (!lim && lead) f[x] = F, vis[x] = true;
	return F;
}
node solve(long long x) {
	if (x < 0) return node();
	tot = 0;
	while (x) {
		num[++tot] = x % 10;
		x /= 10;
	}
	return calc(tot, 1, 0);
}
long long l, r;
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> l >> r;
	std::cout << solve(r) - solve(l - 1) << std::endl;
	return 0;
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9888551.html