[HDU3709]Balanced Number

平衡数计数算法

最新推荐文章于 2024-01-20 17:53:37 发布

转载最新推荐文章于 2024-01-20 17:53:37 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6127145.html

本文介绍了一种计算特定范围内平衡数数量的算法。平衡数是指一个非负整数，通过在某个数字位置放置支点，使得支点左右两侧的力矩相等。文章详细解释了如何使用动态规划的方法来解决这一问题，并提供了完整的C++代码实现。

[HDU3709]Balanced Number

试题描述

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

输入

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10¹⁸).

输出

For each case, print the number of balanced numbers in the range [x, y] in a line.

输入示例

2
0 9
7604 24324

输出示例

10
897

数据规模及约定

见“输入”

题解

令 f[k][i][j][s] 表示考虑数的前 i 位，最高位为 j，支点在位置 k，支点右力矩 - 左力矩 = s 的数有多少个。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
#define LL long long

LL read() {
	LL x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 20
#define maxs 1800
LL f[maxn][maxn][10][maxs];

int num[maxn];
LL sum(LL x) {
	if(!x) return 1;
	int cnt = 0; LL tx = x;
	while(x) num[++cnt] = x % 10, x /= 10;
	LL ans = 0;
	for(int i = cnt - 1; i; i--)
		for(int k = 1; k <= i; k++)
			for(int j = 1; j <= 9; j++) ans += f[k][i][j][0];
	for(int i = cnt; i; i--) {
		for(int k = cnt; k; k--) {
			int s = 0;
			for(int x = cnt; x > i; x--) s += (x - k) * num[x];
			if(s < 0 || s >= maxs) continue;
			for(int j = i < cnt ? 0 : 1; j < num[i]; j++) {
				ans += f[k][i][j][s];
//				if(!j && !s && i > 1) ans--;
			}
		}
	}
	for(int k = 1; k <= cnt; k++) {
		int s = 0;
		for(int x = 1; x <= cnt; x++) s += (x - k) * num[x];
		if(!s){ ans++; break; }
	}
	ans++;
	return ans;
}

int main() {
	for(int j = 0; j <= 9; j++) f[1][1][j][0] = 1;
	for(int k = 2; k < maxn; k++)
		for(int j = 0; j <= 9; j++) f[k][1][j][(k-1)*j] = 1;
	for(int k = 1; k < maxn; k++)
		for(int i = 1; i < maxn - 1; i++)
			for(int j = 0; j <= 9; j++)
				for(int s = 0; s < maxs; s++) if(f[k][i][j][s]) {
					for(int x = 0; x <= 9 && s + (k - i - 1) * x >= 0; x++)
						if(s + (k - i - 1) * x < maxs) f[k][i+1][x][s+(k-i-1)*x] += f[k][i][j][s];
//					printf("%d %d %d %d: %lld\n", k, i, j, s, f[k][i][j][s]);
				}
	int T = read();
	while(T--) {
		LL l = read(), r = read();
		LL ans = sum(r); if(l) ans -= sum(l - 1);
		printf("%lld\n", ans);
	}
	
	return 0;
}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/6127145.html