[河南省ACM省赛-第五届] 最强DE 战斗力 （nyoj 541）

题解链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=541

几天前百度题解后用数学知识AC的，后来大牛说这是一道动态规划题。

网上的数学解题链接：http://blog.csdn.net/x314542916/article/details/8204583

d(i) = max{d(j)*d(n-j) | 1<= j <=n/2};

用Java写比较简单

import java.math.BigInteger;
import java.util.Scanner;

public class Main{

final int N = 1010;
BigInteger d[] = new BigInteger
;
Scanner cin = new Scanner(System.in);

public static void main(String[] args) {
int t, n;
t = cin.nextInt();
for(int i=0; i<N; i++)
d[i] = new BigInteger("0");
for(int i= 1; i<=N-2; i++){
d[i] = BigInteger.valueOf(i);
for(int j=1; j<=i/2; j++){
d[i] = d[i].max(d[j].multiply(d[i-j]));
}
}
while(t-- > 0){
n = cin.nextInt();
System.out.println(d
.toString());
}
}
}

用数学知识的解题代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x7fffffff
#define N 10000
#define MOD 10

int ans
;

void mutify(int n)
{
if(n <= 0)return;
for(int i=0; i<n; i++){
int c = 0;
for(int k=0; k<N; k++){
ans[k] = ans[k]*3+c;
c = ans[k]/MOD;
if(ans[k] >= MOD)
ans[k] %= MOD;
}
}
}

void muti(int n){
for(int i=0; i<N; i++)
ans[i] = ans[i]*n;
int c = 0;
for(int i=0; i<N; i++){
ans[i] = ans[i]+c;
c = ans[i]/MOD;
ans[i] %= MOD;
}
}

int main()
{
int t, m, n;
cin>>t;
while(t--) {
memset(ans, 0, sizeof(ans));
ans[0] =1;
cin>>m;
if(m%3 != 1){
mutify(m/3);
}else {// if(m%3 == 1)
mutify(m/3-1);
if(m > 1) muti(4);
}
if(m%3 == 2){
muti(2);
}
int k = N-1;
while(ans[k] == 0) k--;
for(int i=k; i>=0; i--){
cout<<ans[i];
}
cout<<endl;

}
return 0;
}