洛谷 ——2925干草出售Hay For Sale
题目描述
Farmer John suffered a terrible loss when giant Australian cockroaches ate the entirety of his hay inventory, leaving him with nothing to feed the cows. He hitched up his wagon with capacity C (1 <= C <= 50,000) cubic units and sauntered over to Farmer Don's to get some hay before the cows miss a meal.
Farmer Don had a wide variety of H (1 <= H <= 5,000) hay bales for sale, each with its own volume (1 <= V_i <= C). Bales of hay, you know, are somewhat flexible and can be jammed into the oddest of spaces in a wagon.
FJ carefully evaluates the volumes so that he can figure out the largest amount of hay he can purchase for his cows.
Given the volume constraint and a list of bales to buy, what is the greatest volume of hay FJ can purchase? He can't purchase partial bales, of course. Each input line (after the first) lists a single bale FJ can buy.
约翰遭受了重大的损失:蟑螂吃掉了他所有的干草,留下一群饥饿的牛.他乘着容量为C(1≤C≤50000)个单位的马车,去顿因家买一些干草. 顿因有H(1≤H≤5000)包干草,每一包都有它的体积Vi(l≤Vi≤C).约翰只能整包购买,
他最多可以运回多少体积的干草呢?
输入输出格式
输入格式:
Line 1: Two space-separated integers: C and H
- Lines 2..H+1: Each line describes the volume of a single bale: V_i
输出格式:
- Line 1: A single integer which is the greatest volume of hay FJ can purchase given the list of bales for sale and constraints.
输入输出样例
7 3 2 6 5
7
说明
The wagon holds 7 volumetric units; three bales are offered for sale with volumes of 2, 6, and 5 units, respectively.
Buying the two smaller bales fills the wagon.
代码
#include <algorithm> #include <cstdio> using namespace std; ]; ]; int main() { scanf("%d%d",&v,&n); ;i<=n;i++) scanf("%d",&w[i]); ;i<=n;i++) for(int m=v;m>=w[i];m--) if(f[m-w[i]]+w[i]>f[m]) f[m]=f[m-w[i]]+w[i]; printf("%d",f[v]); ; }
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