rt

题解:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

const int MAXN = 85, Mod = 10000; //高精四位压缩大法好 
int n, m;
int ar[MAXN];

struct HP {
	int p[505], len;
	HP() {
		memset(p, 0, sizeof p);
		len = 0;
	} //这是构造函数,用于直接创建一个高精度变量 
	void print() {
		printf("%d", p[len]);  
        for (int i = len - 1; i > 0; i--) {  
            if (p[i] == 0) {
				printf("0000"); 
				continue;
			}
            for (int k = 10; k * p[i] < Mod; k *= 10) 
				printf("0");
            printf("%d", p[i]);
        }
	} //四位压缩的输出 
} f[MAXN][MAXN], base[MAXN], ans;

HP operator + (const HP &a, const HP &b) {
	HP c; c.len = max(a.len, b.len); int x = 0;
	for (int i = 1; i <= c.len; i++) {
		c.p[i] = a.p[i] + b.p[i] + x;
		x = c.p[i] / Mod;
		c.p[i] %= Mod;
	}
	if (x > 0)
		c.p[++c.len] = x;
	return c;
} //高精+高精 

HP operator * (const HP &a, const int &b) {
	HP c; c.len = a.len; int x = 0;
	for (int i = 1; i <= c.len; i++) {
		c.p[i] = a.p[i] * b + x;
		x = c.p[i] / Mod;
		c.p[i] %= Mod;
	}
	while (x > 0)
		c.p[++c.len] = x % Mod, x /= Mod;
	return c;
} //高精*单精 

HP max(const HP &a, const HP &b) {
	if (a.len > b.len)
		return a;
	else if (a.len < b.len)
		return b;
	for (int i = a.len; i > 0; i--)
		if (a.p[i] > b.p[i])
			return a;
		else if (a.p[i] < b.p[i])
			return b;
	return a;
} //比较取最大值 

void BaseTwo() {
	base[0].p[1] = 1, base[0].len = 1;
	for (int i = 1; i <= m + 2; i++){ //这里是m! m! m! 我TM写成n调了n年... 
		base[i] = base[i - 1] * 2;
	}
} //预处理出2的幂 

int main(void) {
	scanf("%d%d", &n, &m);
	BaseTwo();
	while (n--) {
		memset(f, 0, sizeof f);
		for (int i = 1; i <= m; i++)
			scanf("%d", &ar[i]);
		for (int i = 1; i <= m; i++)
			for (int j = m; j >= i; j--) { //因为终值是小区间,DP自然就从大区间开始 
				f[i][j] = max(f[i][j], f[i - 1][j] + base[m - j + i - 1] * ar[i - 1]); 
				f[i][j] = max(f[i][j], f[i][j + 1] + base[m - j + i - 1] * ar[j + 1]);
			} //用结构体重载运算符写起来比较自然 
		HP Max;
		for (int i = 1; i <= m; i++)
			Max = max(Max, f[i][i] + base[m] * ar[i]);
		ans = ans + Max; //记录到总答案中 
	}
	ans.print(); //输出 
	return 0;
}

他的:

#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int MAXN = 85, Mod = 10000;
int n, m;
int ar[MAXN];
struct HP
{
	int p[505], len;
	HP()
    {
		memset(p, 0, sizeof p);
		len = 0;
	}
	void print()
    {
		printf("%d", p[len]);  
        for (int i = len - 1; i > 0; i--)
        {  
            if (p[i] == 0)
            {
				printf("0000"); 
				continue;
			}
            for (int k = 10; k * p[i] < Mod; k *= 10)
            { 
				printf("0");
            }
            printf("%d", p[i]);
        }
	}
} f[MAXN][MAXN], base[MAXN], ans;
HP operator + (const HP &a, const HP &b)
{
	HP c; c.len = max(a.len, b.len); int x = 0;
	for (int i = 1; i <= c.len; i++)
    {
		c.p[i] = a.p[i] + b.p[i] + x;
		x = c.p[i] / Mod;
		c.p[i] %= Mod;
	}
	if (x > 0)
		c.p[++c.len] = x;
	return c;
}
HP operator * (const HP &a, const int &b)
{
	HP c; c.len = a.len; int x = 0;
	for (int i = 1; i <= c.len; i++)
    {
		c.p[i] = a.p[i] * b + x;
		x = c.p[i] / Mod;
		c.p[i] %= Mod;
	}
	while (x > 0)
    {
		c.p[++c.len] = x % Mod, x /= Mod;
    }
	return c;
}
HP max(const HP &a, const HP &b)
{
	if (a.len > b.len)
    {
		return a;
    }
	else if (a.len < b.len)
    {
		return b;
    }
	for (int i = a.len; i > 0; i--)
    {
		if (a.p[i] > b.p[i])
        {
			return a;
        }
		else if (a.p[i] < b.p[i])
        {
			return b;
        }
    }
	return a;
} 
void BaseTwo()
{
	base[0].p[1] = 1, base[0].len = 1;
	for (int i = 1; i <= m + 2; i++)
    {
		base[i] = base[i - 1] * 2;
	}
}
int main(void)
{
	scanf("%d%d", &n, &m);
	BaseTwo();
	while (n--)
    {
		memset(f, 0, sizeof f);
		for (int i = 1; i <= m; i++)
        {
			scanf("%d", &ar[i]);
        }
		for (int i = 1; i <= m; i++)
        {
			for (int j = m; j >= i; j--)
            {
				f[i][j] = max(f[i][j], f[i - 1][j] + base[m - j + i - 1] * ar[i - 1]); 
				f[i][j] = max(f[i][j], f[i][j + 1] + base[m - j + i - 1] * ar[j + 1]);
			}
        }
		HP Max;
		for (int i = 1; i <= m; i++)
        {
			Max = max(Max, f[i][i] + base[m] * ar[i]);
        }
		ans = ans + Max;
	}
	ans.print();
	return 0;
}

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