HDU 1502

来源：互联网发布：国家网络信息安全中心编辑：程序博客网时间：2024/06/11 09:02

http://acm.hdu.edu.cn/showproblem.php?pid=1502

题意：给你n个A，n个B，n个C，求出满足条件if c is a prefix of w , then A(c)>= B(c) >= C(c) .的排列的个数。

题一看完就想到了记忆化搜索，觉得挺简单的，就直接开始写了，等写完了才发现，答案实在是太大了。必须要用大整数的加法才行，然后又改成了大整数加法，结果看似对了，结果一提交居然超时了。想了一下又该成了打表的形式，这才终于过了。不过内存占用简直大得不敢直视啊。

后来又去看了一下别人的解题报告，用递推式dp[i][j][k] = dp[i-1][j][k]+dp[i][j-1][k]+dp[i][j][k-1]就可得出答案;

以下为两种解法的代码：

#include <cstdio>#include <cstring>#include <iostream>#include <queue>#include <algorithm>#include <map>#include <vector>#include <stack>using namespace std;#define M 1000005#define ll long long#define int64 __int64int n;struct node{    int fg;    char num[105];    node()    {        memset(num , 0 , sizeof num);    }}dp[65][65][65];void Cal(char * a , char * b){    for (int i = 100 ; i > 0 ; i--)    {        a[i] += b[i];        if (a[i] > 9)        {            a[i-1] += a[i]/10;            a[i] %= 10;        }    }}node Dfs(int a , int b , int c){    node ret;    ret.fg = 1;    if (a == 60 && b == 60 && c == 60)    {        ret.num[100] = 1;        return ret;    }    if (dp[a][b][c].fg)return dp[a][b][c];    if (a < 60)Cal(ret.num,Dfs(a+1,b,c).num);    if (b < 60 && b < a)Cal(ret.num,Dfs(a,b+1,c).num);    if (c < 60 && c < a && c < b)Cal(ret.num,Dfs(a,b,c+1).num);    return dp[a][b][c] = ret;}int main(){    memset(dp , 0 , sizeof dp);    Dfs(0,0,0);    while (~scanf("%d",&n))    {        node ans = dp[60-n][60-n][60-n];        int i = 0;        while (ans.num[i] == 0 && i < 100)i++;        while (i <= 100)printf("%d",ans.num[i]),i++;        printf("\n\n");    }    return 0;}

#include <cstdio>#include <cstring>#include <iostream>#include <queue>#include <algorithm>#include <map>#include <vector>#include <stack>using namespace std;#define M 1000005#define ll long long#define int64 __int64int n;char dp[65][65][65][105];void Cal(char * a , char * b){    for (int i = 100 ; i > 0 ; i--)    {        a[i] += b[i];        if (a[i] > 9)        {            a[i-1] += a[i]/10;            a[i] %= 10;        }    }}void Init(){    int i , j , k;    memset(dp , 0 , sizeof dp);    dp[0][0][0][100] = 1;    for (i = 1 ; i < 61 ; i++)    {        for (j = 0 ; j <= i ; j++)        {            for (k = 0 ; k <= j ; k++)            {                Cal(dp[i][j][k],dp[i-1][j][k]);                if (j > 0)Cal(dp[i][j][k],dp[i][j-1][k]);                if (k > 0)Cal(dp[i][j][k],dp[i][j][k-1]);            }        }    }}int main(){    Init();    while (~scanf("%d",&n))    {        int i = 0;        dp[0][0][0][100] = 0;        while (dp[n][n][n][i] == 0 && i < 100)i++;        while (i <= 100)printf("%d",dp[n][n][n][i++]);        printf("\n\n");    }    return 0;}