Boxes and Stones
Paul and Carole like to play a game with
S stones and B boxes numbered from 1 to
B. Beforebeginning the game they arbitrarily distribute the
S stones among the boxes from 1 to
B - 1, leavingbox B empty. The game then proceeds by rounds. At each round, first Paul chooses a subset
P of the stones that are in the boxes; he may choose as many stones as he wants from as many boxes as he wants, or he may choose no stones at all, in which case
P is empty. Then, Carole decides what to donext: she can either
promote the subset P and
discard the remaining stones (that is, those stones notchosen by Paul in the first step); or she may
discard the subset P and
promote the remaining stones.
To promote a given subset means to take each stone in this subset and move it to the box with thenext number in sequence, so that if there was a stone in this subset inside box
b, it is moved to boxb + 1. To
discard a given subset means to remove every stone in this subset from its corresponding box, so that those stones are not used in the game for the remaining rounds. The figure below shows an example of the first two rounds of a
game.
Paul and Carole play until at least one stone reaches box number
B, in which case Paul wins the game, or until there are no more stones left in the boxes, in which case Carole wins the game. Paulis a very rational player, but Carole is a worthy rival because she is not only extremely good
at this game, but also quite lucky. We would like to know who is the best player, but before that we must first understand how the outcome of a game depends on the initial distribution of the stones. In particular,we would like to know in how many ways the
S stones can initially be distributed among the first B - 1 boxes so that Carole can be certain that she can win the game if she plays optimally, even if Paul never makes a mistake.
Input
Each test case is described using one line. The line contains two integers
S (1
S200)
and B(2B100),
representing respectively the number of stones and the number of boxes in the game.
Output
For each test case output a line with an integer representing the number of ways in which the
S stonesmay be distributed among the first
B - 1 boxes so that Carole is certain that she can win the game.Because this number can be very large, you are required to output the remainder of dividing it by109 + 7.
Sample Input
2 3 8 4 42 42
Sample Output
2 0 498467348
题目大意：P和C做游戏。有S个石头和B个盒子，他们把S个石头随意的放在1～B-1这些盒子中，P開始从1～B-1中随意的挑选一些石头。C能够决定把P选的石头丢掉让剩下的石头都往右移一格，或者把P选的石头都往右移一个。其它的都丢掉。一旦有一个石头到B盒子中。则P胜。若没有石头了则C胜，问有多少种初始的摆放状态为C的必胜态
题目分析：设dp[i][j][k]为到第i个盒子。用了j个石头，C做完决定后第i个盒子里的石头数为k的必胜态个数。离线计算出全部情况。O(1)查询，先分析P的最优策略，P的目标是尽量多的让石子右移。但是选择权在C手上，因此P的最优策略是尽量一半一半的取，假设在这样的情况下P还是输，那么就是C的必胜态了。dp[i][j][k]状态由三部分转移
dp[i][j][k] = dp[i][j - 1][k - 1] + dp[i - 1][j][2 * k] + dp[i - 1][j][2 * k + 1]
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
int const MOD = 1e9 + 7;
int const MAX = 105;
ll dp[MAX][2 * MAX][4 * MAX];
void pre()
{
dp[0][0][0] = 1;
for(int i = 1; i <= 100; i++)
for(int j = 0; j <= 200; j++)
for(int k = 0; k <= 200; k++)
dp[i][j][k] = (dp[i][j - 1][k - 1] + dp[i - 1][j][k * 2] + dp[i - 1][j][k * 2 + 1]) % MOD;
}
int main()
{
pre();
int s, b;
while(scanf("%d %d", &s, &b) != EOF)
printf("%lld\n", dp[b][s][0]);
}