POJ 3601：Tower of Hanoi

Tower of Hanoi

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 1903		Accepted: 651

Description

The Tower of Hanoi is a puzzle consisting of three pegs and a number of disks of different sizes which can slide onto any peg. The puzzle starts with the disks neatly stacked in order of size on one peg, the smallest at the top, thus making a conical shape.
The objective of the puzzle is to move the entire stack to another peg, obeying the following rules:

Only one disk may be moved at a time.
Each move consists of taking the upper disk from one of the pegs and sliding it onto another peg, on top of the other disks that may already be present on that peg.
No disk may be placed on top of a smaller disk.
For n disks, it is a well-known result that the optimal solution takes 2n − 1 moves.

To complicate the puzzle a little, we allow multiple disks to be of the same size. Moreover, equisized disks are mutually distinguishable. Their ordering at the beginning should be preserved at the end, though it may be disturbed during the process of solving
the puzzle.

Given the number of disks of each size, compute the number of moves that the optimal solution takes.

Input

The input contains multiple test cases. Each test case consists of two lines. The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 106). The second lines contains n integers a1, a2,
…, an (1 ≤ a1,a2, …, an ≤ 105). For each 1 ≤ i ≤ n, there are ai disks of size i. The input ends where EOF is met.

Output

For each test case, print the answer modulo m on a separate line.

Sample Input

1 1000
2
5 1000
1 1 1 1 1
5 1000
2 2 2 2 2
5 1000
1 2 1 2 1

Sample Output

3
31
123
41

Source

PKU Campus 2008 (POJ Founder Monthly Contest – 2008.05.10), xfxyjwf

题意还是求汉诺塔的最少挪动次数，不同的是这次有相同大小的盘子了。

当第k大盘子只有一个的时候很好办，想象一下，相同大小的盘子当成一个盘子，放一次放倒了，经过这一次，正好放正回来了。

当第k大盘子有多个的时候，挪动的时候假设k-1个盘子在柱子上已经放好了，那么相同大小的盘子要顺序一致，必须腾出两个柱子来挪，这部分是2*val[k]，之前的那k-1个盘子再都使用当成一个盘子的方法，颠一次倒一次，正好放正了。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
using namespace std;
typedef long long ll;

#define eps 1e-8
#define INF 1<<25
#define repp(i, n, m) for (int i = n; i <= m; i++)
#define rep(i, n, m) for (int i = n; i < m; i++)
#define sa(n) scanf("%d", &(n))
#define mp make_pair
#define ff first
#define ss second
#define pb push_back

const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
const double PI = acos(-1.0);

int n, m;
int val[maxn], cnt[maxn], dp[maxn];

void solve()
{
int i, j, k;
repp(i, 1, n)
{
sa(val[i]);
}
repp(i, 1, n)
{
cnt[i] = ((cnt[i - 1] << 1) + val[i]) % m;
}
dp[1] = 2 * val[1] - 1;
repp(i, 2, n)
{
if (val[i] == 1)
{
dp[i] = cnt[i];
}
else
{
dp[i] = (2 * cnt[i - 1] + 2 * val[i] + dp[i - 1]) % m;
}
}
printf("%d\n", dp
);
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("i.txt", "r", stdin);
freopen("o.txt", "w", stdout);
#endif
while (scanf("%d%d", &n, &m) != EOF)
{
solve();
}
return 0;
}