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# include <bits/stdc++.h>
// https://www.acwing.com/solution/content/39331/
using namespace std ;
const int N = 1010 ;
int n , m ;
int v [ N ] , w [ N ] ;
int f [ N ] [ N ] ;
/**
测试用例:
80 2
18 10
30 20
体积大小n=80, 物品个数: m=2
第一个物品 体积w[1]=18, 产生的价值v[1]=10
第二个物品 体积w[2]=30, 产生的价值v[2]=20
选择方法, 第一个物品选择1个, 第二个物品选择2个, 共10+20*2=50
*/
//完全背包问题
int main ( ) {
cin > > m > > n ; //n是物品种类, m是背包上限, 本题是钱数
for ( int i = 1 ; i < = n ; i + + ) cin > > w [ i ] > > v [ i ] ; //注意这里的顺序,背包上限是钱,分清价格是重量,重量是价值,这里是反的
for ( int i = 1 ; i < = n ; i + + ) //考虑前i个物品
for ( int j = 1 ; j < = m ; j + + ) //从1开始到m去考虑重量下的最优解
//对比 LanQiao12ShengSai/BC4_ErWei_DP.cpp 01背包
if ( j > = w [ i ] ) f [ i ] [ j ] = max ( f [ i ] [ j ] , f [ i ] [ j - w [ i ] ] + v [ i ] ) ; //还可以在i中商品中选择, 注意01背包是f[i-1][j]
else f [ i ] [ j ] = f [ i - 1 ] [ j ] ; //不要这个物品
//输出结果
cout < < f [ n ] [ m ] < < endl ;
return 0 ;
}