WOJ1005 - Holding Animals

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诺亚方舟的故事中，动物太多无法全部容纳。每种动物都有一个代表上帝喜爱程度的分数。诺亚需要决定让哪些动物上船以最大化总分数，但他不会编程。这是一个简单的01背包问题，输入包含动物种类数量、大小和分数，以及方舟的容量。输出应为能获得的最大分数。

The total available floor space on the ark would have been over 100,000 square feet, which would be more floor space than in 20 standard sized
basketball courts. But in our story, we assume that the land dwelling air breathing animals is so many that the Ark can't contain all of them.

We know that the size of each kind of animal is different and the God has his own favorite. The God assigns each kind of animal a point to
show his favorite, and then lists points of every kind of animal to Noah. Noah must let some kinds of animals into the Ark and refuse
the others to maximize the total points. Noah cries for he doesn't know computer programming.

It 's an easy problem for you,right? So,rescue Noah!

输入格式

There will be multiple test cases. For each test case,the first line contains an integer n(n<=100) representing the number of species of the land
dwelling air breathing animals all over the world.
In the next n lines,there will be two integers in each line,separated by a single space,where the first integer shows the size of a kind of animal
and the second integer shows the point. The size and the point for all the animals will not exceed 10000.
The last line contains an integer s(s <= 100,000) indicating the size of Noah's Ark.

输出格式

For each test case,you should output a line that contains a single integer to describe the maximal point Noah can get.

样例输入

样例输出

50
40

裸地01背包问题

简易公式：f[i]=max(f[i],f[i-w[j]]+v[j])

#include <stdio.h>  
int main(){
	int n,smax,i,j;
	int s[10001],p[10001],f[100010];
	while(scanf("%d",&n)==1){
		for(i=1;i<=n;i++){
			scanf("%d %d",&s[i],&p[i]);
		}
		scanf("%d",&smax);
		for(j=0;j<=smax;j++){
			f[j]=0;
		}
		for(i=1;i<=n;i++){
			for(j=smax;j>=1;j--){
				if(j-s[i]>=0)
				f[j]=f[j]>(f[j-s[i]]+p[i])?f[j]:(f[j-s[i]]+p[i]); 
			}
		}
		printf("%d\n",f[smax]);
	}
	return 0;
}