BOJ 953 flower

//============================================================================// Name        : 953.cpp// Author      : wly// Version     :// Copyright   : flower// Description : 曾经写过的一道题，4维的dp//令dp[i][j][k][m]存放i个A种花，j个B种花，k个C种花，且以第m种花作为最后一朵//dp方程：//dp[i][j][k][LAST_IS_A] = dp[i-1][j][k][LAST_IS_B] + dp[i-1][j][k][LAST_IS_C]//dp[i][j][k][LAST_IS_B] = dp[i][j-1][k][LAST_IS_A] + dp[i][j-1][k][LAST_IS_C]//dp[i][j][k][LAST_IS_C] = dp[i][j][k-1][LAST_IS_A] + dp[i][j][k-1][LAST_IS_B]//可以看出：//dp[a][b][c][LAST_IS_A] + dp[a][b][c][LAST_IS_B] + dp[a][b][c][LAST_IS_C]即为最终结果//注意://处理边缘情况//dp[0][0][0][LAST_IS_A] = dp[0][0][0][LAST_IS_B] = dp[0][0][0][LAST_IS_C] = 1;//dp[1][0][0][LAST_IS_A] = dp[0][1][0][LAST_IS_B] = dp[0][0][1][LAST_IS_C] = 1;//预处理:先将所有情况直接算出来,之后根据读入的数据直接输出就好,避免TLE//============================================================================#include <iostream>#include <cstdlib>#include <cstdio>#include <cmath>#define LAST_IS_A 0#define LAST_IS_B 1#define LAST_IS_C 2#define MOD 99999999int dp[65][65][65][3];using namespace std;int main(void) {int a, b, c;memset(dp, 0, sizeof(dp));dp[0][0][0][LAST_IS_A] = dp[0][0][0][LAST_IS_B] = dp[0][0][0][LAST_IS_C]= 1;dp[1][0][0][LAST_IS_A] = dp[0][1][0][LAST_IS_B] = dp[0][0][1][LAST_IS_C]= 1;for (int i = 0; i <= 64; i++)for (int j = 0; j <= 64; j++)for (int k = 0; k <= 64; k++) {if (i != 0 && (j != 0 || k != 0))dp[i][j][k][LAST_IS_A] = (dp[i - 1][j][k][LAST_IS_B] + dp[i- 1][j][k][LAST_IS_C]) % MOD;if (j != 0 && (i != 0 || k != 0))dp[i][j][k][LAST_IS_B] = (dp[i][j - 1][k][LAST_IS_A]+ dp[i][j - 1][k][LAST_IS_C]) % MOD;if (k != 0 && (i != 0 || j != 0))dp[i][j][k][LAST_IS_C] = (dp[i][j][k - 1][LAST_IS_A]+ dp[i][j][k - 1][LAST_IS_B]) % MOD;}int tc = 1;while (~scanf("%d%d%d", &a, &b, &c)) {printf("Case %d:\n%d\n", tc++, (dp[a][b][c][LAST_IS_A]+ dp[a][b][c][LAST_IS_B] + dp[a][b][c][LAST_IS_C]) % MOD);}}