#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
int a[101],c[101];
int num[100001],f[100001];
int main()
{
int n,m;
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
int sum=0;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
scanf("%d",&c[i]);
memset(f,0,sizeof(f));
f[0]=1;
for(int i=1;i<=n;i++)
{
memset(num,0,sizeof(num));
for(int j=a[i];j<=m;j++)
{
if(!f[j]&&f[j-a[i]]&&num[j-a[i]]<c[i])
{
num[j]=num[j-a[i]]+1;
sum++;
f[j]=1;
}
}
}
printf("%d\n",sum);
}
return 0;
}

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
int d[50000],v[12],w[12];//数组开大了会超时
int main()
{

    int n;
    //freopen("out.txt","w",stdout);
    scanf("%d",&n);
    for(int l=1;l<=n;l++)
    {
        int c,k;
        int m;
        scanf("%d %d",&c,&k);
        scanf("%d",&m);

        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&v[i],&w[i]);
            v[i]/=1000;
        }
        while(k--)
        {
             int t=c/1000;//因为是1000的倍数 所以 都除以1000 优化一下子
        memset(d,0,sizeof(d));
        for(int i=1;i<=m;i++)
         for(int j=0;j<=t;j++)
         {   if(j>=v[i])
             if(d[j]<(d[j-v[i]]+w[i]))
            {
                d[j]=d[j-v[i]]+w[i];
            }
        }
         c=c+d[t];
        }
         printf("%d\n",c);
    }
    return 0;
}

 int cmp(const void *a,const void *b);

 其中返回值必须是int,两个参数的类型必须都是const void *, 这样你可以对他进行强制转换成任何其他类型
 有点想C#中的box  哈~~
 假设是对int排序的话,如果是升序,那么就是如果a比b大cmp返回一个正值,小则负值,相等返回0
 当然当你对具体的数组进行排序的话 cmp里面就要做具体的处理
 
        2。  七种常见排法！！！
 一、对int类型数组排序
 int num[100];
 Sample:

 int cmp ( const void *a , const void *b )
 {
 return *(int *)a - *(int *)b;
 }
 qsort(num,100,sizeof(num[0]),cmp);

 二、对char类型数组排序（同int类型）
 char word[100];
 Sample:

 int cmp( const void *a , const void *b )
 {
 return *(char *)a - *(char *)b;
 }
 qsort(word,100,sizeof(word[0]),cmp);

 三、对double类型数组排序（特别要注意）

 double in[100];
 int cmp( const void *a , const void *b )
 {
 return *(double *)a > *(double *)b ? 1 : -1;
 }
 qsort(in,100,sizeof(in[0]),cmp)；

 四、对结构体一级排序
 

struct In
    {
    double data;
    int other;
    }s[100]
    //按照data的值从小到大将结构体排序,关于结构体内的排序关键数据data的类型可以很多种，参考上面的例子写
    int cmp( const void *a ,const void *b)
    {
    return (*(In *)a)->data > (*(In *)b)->data ? 1 : -1;
    }
    qsort(s,100,sizeof(s[0]),cmp)

 五、对结构体二级排序
 

struct In
{
int x;
int y;
}s[100];//按照x从小到大排序，当x相等时按照y从大到小排序
int cmp( const void *a , const void *b )
{
struct In *c = (In *)a;
struct In *d = (In *)b;
if(c->x != d->x) return c->x - d->x;
else return d->y - c->y;
}
qsort(s,100,sizeof(s[0]),cmp);

 六、对字符串进行排序

struct In
    {
    int data;
    char str[100];
    }s[100];
    //按照结构体中字符串str的字典顺序排序
    int cmp ( const void *a , const void *b )
    {
    return strcmp( (*(In *)a)->str , (*(In *)b)->str );
    }
    qsort(s,100,sizeof(s[0]),cmp);

 七、计算几何中求凸包的cmp

 int cmp(const void *a,const void *b) //重点cmp函数，把除了1点外的所有点，旋转角度排序
 {
 struct point *c=(point *)a;
 struct point *d=(point *)b;
 if( calc(*c,*d,p[1]) < 0) return 1;
 else if( !calc(*c,*d,p[1]) && dis(c->x,c->y,p[1].x,p[1].y) < dis(d->x,d->y,p[1].x,p[1].y)) //如果在一条直线上，则把远的放在前面
 return 1;
 else return -1;
 }

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 78   Accepted Submission(s) : 47

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Problem Description

Input

Output

Sample Input

2 3
1 2
1 2
3 5
0 3
0 3
0 3

Sample Output

2
12

Author

Source

 1#include<stdio.h>
 2int main()
 3{
 4    int i,n,m,j,k,a[101],b[101],c1[101],c2[101];
 5    while(scanf("%d %d",&n,&m)!=EOF)
 6    {
 7        for(i=0;i<n;i++)
 8        scanf("%d %d",&a[i],&b[i]);
 9        for(i=0;i<=m;i++)
10        {
11            c1[i]=0;
12            c2[i]=0;
13        }
14        c1[0]=1;//第一次结束以后 k的就被赋值我为1了 可其余还为0 就算c1[j]中被用到还是没效果 只有1的c1[j]才会有效果
15        for(i=0;i<n;i++)
16        {
17            for(j=0;j<=m;j++)
18         for(k=a[i];k+j<=m&&k<=b[i];k++)
19               c2[j+k]+=c1[j];
20           for(j=0;j<=m;j++)
21        {
22            c1[j]=c2[j];
23            c2[j]=0;
24        }
25        }
26        printf("%d\n",c1[m]);
27    }
28    return 0;
29}

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 132   Accepted Submission(s) : 69

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Problem Description

We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
“Oh, God! How terrible! ”



Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

Input

Output

Sample Input

1 1 3
0 0 0

Sample Output

4

Author

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,k,a,b,c,d[10001];
    while(scanf("%d %d %d",&a,&b,&c)!=EOF)
    {
        if(0==a&&0==b&&0==c)
        break;
        memset(d,0,sizeof(d));
          for(j=0;j<=a;j++)
          for(k=0;k+j<=a+2*b;k=k+2)
             d[k+j]++;
             for(j=0;j<=a+2*b&&d[j];j++)//&&a[j]是排除从0到a+2b之内不存在的项
             for(k=0;k+j<=a+2*b+5*c;k=k+5)
              d[k+j]++;
              for(i=0;i<=a+2*b+5*c+1;i++)//+1是因为如果情况中的全满足 那么大于范围的第一个就是所求
             {
                 if(d[i]==0)
                 {
                     printf("%d\n",i);
                     break;
                 }
             }
    }
    return 0;
}