codeforces Gym 100500C C. ICPC Giveaways 暴力

Problem C. ICPC Giveaways
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100500/attachments

Description

During the preparation for the ICPC contest, the organizers prepare bags full of giveaways for the contestants. Each bag usually contains an MP3 Player, a Sim Card, a USB HUB, a USB Flash Drive, ... etc. A problem happened during the delivery of the components of the bags, so not every component was delivered completely to the organizers. For example the organizers ordered 10 items of 4 different types, and what was delivered was 7, 6, 8, 9 from each type respectively. The organizers decided to form bags anyway, but they have to abide by 2 rules. All formed bags should have exactly the same items, and no bag should contain 2 items of the same type (either 1 or 0). Knowing that each item has an amusement value (for sure an MP3 Player is much more amusing than a Sim Card), the organizers decided to get the max possible total amusement. The total amusement is the amusement value of a single bag multiplied by the number of bags. Note that not every contestant should receive a bag. The amusement value of each item type is calculated using this equation:(i × i) mod C where i is an integer that represents the item type, and C is a value that will be given as an input. Please help the ICPC organizers to determine what the maximum total amusement is.

Input

T is the number of test cases. For each test case there will be 3 integers M,N and C, where M is the number of items, N is the total number of types, and C is as described above then M integer representing the type of each item. 1 ≤ T ≤ 100 1 ≤ M ≤ 10, 000 1 ≤ N ≤ 10, 000 1 ≤ C ≤ 10, 000 1 ≤ itemi ≤ N

Output

For each test case print a single line containing: Case_x:_y x is the case number starting from 1. y is the required answer. Replace the underscores with spaces.

Sample Input

1 10 3 9 1 1 2 2 1 1 2 3 1 2

Sample Output

Case 1: 20

HINT

题意

要求从已有的物品中选取若干个，组成若干个一模一样的bag，要求每个bag中同类物品只有一个，最后选手满意度=bag数目*每个bag满意度

题解：

满意度=sum(value(k))*bag,k满足cnt[k]>=bag，那么只要排序，维护前缀和，然后扫描一遍数组即可

//debug:1.一开始就没有注意到mod c是针对每一种物品的而不是针对总和的

代码：

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <sstream>
10 #include <queue>
11 #include <typeinfo>
12 #include <fstream>
13 #include <map>
14 #include <stack>
15 typedef long long ll;
16 using namespace std;
17 #define test freopen("1.txt","r",stdin)
18 #define maxn 2000001
19 #define mod 10007
20 #define eps 1e-9
21 const int inf=0x3f3f3f3f;
22 const ll infll = 0x3f3f3f3f3f3f3f3fLL;
23 inline int read()
24 {
25     ll x=0,f=1;
26     char ch=getchar();
27     while(ch<‘0‘||ch>‘9‘)
28     {
29         if(ch==‘-‘)f=-1;
30         ch=getchar();
31     }
32     while(ch>=‘0‘&&ch<=‘9‘)
33     {
34         x=x*10+ch-‘0‘;
35         ch=getchar();
36     }
37     return x*f;
38 }
39 inline void out(int x)
40 {
41     if(x>9) out(x/10);
42     putchar(x%10+‘0‘);
43 }
44 //**************************************************************************************
45 struct item{
46     int v;
47     int cnt;
48 }item[10000+100];
49 bool cmp(struct  item a,struct item b)
50 {
51     return a.cnt>b.cnt;
52 }
53 int sum[10000+100];
54 int main()
55 {
56     int t=read();
57     for(int cc=1;cc<=t;cc++)
58     {
59         int m=read(),n=read(),c=read();
60         memset(sum,0,sizeof(sum));
61         for(int i=1;i<=n;i++){
62             item[i].cnt=0;
63             item[i].v=i*i%c;
64         }
65         for(int i=1;i<=m;i++){
66             int type=read();
67             item[type].cnt++;
68         }
69         sort(item+1,item+n+1,cmp);
70         for(int i=1;i<=n;i++){
71             if(i==1) sum[i]=item[i].v;
72             else sum[i]=sum[i-1]+item[i].v;
73         }
74         int maxx=-1;
75         for(int i=1;i<=n;i++){
76             maxx=max(maxx,item[i].cnt*sum[i]);
77         }
78         printf("Case %d: %d\n",cc,maxx);
79     }
80     return 0;
81 }

另外，以c为标准维护前缀和不知道为什么错。。。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
#define test freopen("1.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline int read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)
    {
        if(ch==‘-‘)f=-1;
        ch=getchar();
    }
    while(ch>=‘0‘&&ch<=‘9‘)
    {
        x=x*10+ch-‘0‘;
        ch=getchar();
    }
    return x*f;
}
inline void out(int x)
{
    if(x>9) out(x/10);
    putchar(x%10+‘0‘);
}
//**************************************************************************************
struct item{
    int v;
    int cnt;
}item[10000+100];
bool cmp(struct  item a,struct item b)
{
    return a.cnt>b.cnt;
}
int sum[10000+100];
int main()
{
    int t=read();
    for(int cc=1;cc<=t;cc++)
    {
        int m=read(),n=read(),c=read();
        memset(sum,0,sizeof(sum));
        for(int i=0;i<c;i++){
            item[i].cnt=0;
            item[i].v=i;
        }
        for(int i=1;i<=m;i++){
            int type=read();
            int v=type*type%c;
            item[v].cnt++;
        }
        sort(item,item+c,cmp);
        for(int i=0;i<c;i++){
            if(i==0) sum[i]=item[i].v;
            else sum[i]=sum[i-1]+item[i].v;
        }
        int maxx=-1;
        for(int i=0;i<c;i++){
           // printf("%d\n",item[i].cnt);
            maxx=max(maxx,item[i].cnt*sum[i]);
        }
        printf("Case %d: %d\n",cc,maxx);
    }
    return 0;
}