# 动态规划进阶题目之Charm Bracelet

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

``````4 6
1 4
2 6
3 12
2 7``````

``23``

``````#include<stdio.h>
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int n,m,i,j;
int w[4000],p[4000];
while(scanf("%d%d",&n,&m)!=EOF)
{
int c[20000]={0};//赋零值，下面比较用
for(i=1;i<=n;i++)
scanf("%d%d",&w[i],&p[i]);
for(i=1;i<=n;i++)//循环n件物品
for(j=m;j>=w[i];j--)//循环重量，0到m，不断比较加上第i件物品与不加上第i件物品的价值，值大的赋值给c[j]
c[j]=max(c[j],c[j-w[i]]+p[i]);
printf("%d\n",c[m]);
}
return 0;
}``````

``````#include<stdio.h>

int n,max_w,max_d,i,t;
int w[3403],d[3403];
int f[12881];

int main()
{
scanf("%d%d",&n,&max_w);
for (i=1;i<=n;i++)
scanf("%d%d",&w[i],&d[i]);
for (i=1;i<=n;i++)
for (t=max_w-w[i];t>=0;t--)
if (f[t+w[i]]<f[t]+d[i])
f[t+w[i]]=f[t]+d[i];
for (t=0;t<=max_w;t++)
if (f[t]>max_d)
max_d=f[t];
printf("%d\n",max_d);
return 0;
} ``````