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Description
Alex doesn‘t like boredom. That‘s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let‘s denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings akpoints to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input
The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex‘s sequence.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).
Output
Print a single integer — the maximum number of points that Alex can earn.
Sample Input
2
1 2
2
3
1 2 3
4
9
1 2 1 3 2 2 2 2 3
10
Hint
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
My code：
1 #include<stdio.h> 2 #include<string.h> 3 #include<math.h> 4 #include<algorithm> 5 using namespace std; 6 bool cmp(int x,int y) 7 { 8 return x>y; 9 } 10 11 int main() 12 { 13 int n; 14 long long dp[100005]; 15 int a[100005],b[100005],m[100005]; 16 int i,j; 17 while(scanf("%d",&n)!=EOF) 18 { 19 memset(m,0,sizeof(m)); 20 memset(a,0,sizeof(a)); 21 memset(dp,0,sizeof(dp)); 22 for(i=1;i<=n;i++) 23 { 24 scanf("%d",&a[i]); 25 if(m[a[i]]>0) 26 { 27 m[a[i]]++; 28 a[i]=0;i--;n--; 29 } 30 else 31 { 32 m[a[i]]++; 33 } 34 } 35 /*printf("%d\n",n); 36 for(i=1;i<=n;i++) 37 printf("%d ",a[i]);*/ 38 39 sort(a+1,a+n+1,cmp); 40 dp[0]=0; 41 dp[1]=(long long)a[1]*m[a[1]]; 42 if(a[1]-1==a[2]) 43 dp[2]=(long long)a[2]*m[a[2]]; 44 else 45 dp[2]=dp[1]+(long long)a[2]*m[a[2]]; 46 //printf("%I64d %I64d\n",dp[1],dp[2]); 47 for(i=3;i<=n;i++) 48 { 49 long long maa=0; 50 maa=max(dp[i-2],dp[i-3]); 51 if(a[i-1]-1==a[i]) 52 dp[i]=maa+(long long)a[i]*m[a[i]]; 53 else 54 { 55 dp[i]=(long long)a[i]*m[a[i]]+max(maa,dp[i-1]); 56 } 57 } 58 long long ans=0; 59 for(i=1;i<=n;i++) 60 if(ans<dp[i]) 61 ans=dp[i]; 62 printf("%I64d\n",ans); 63 } 64 return 0; 65 }
别宁家的代码 短小&精悍
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 #define LL __int64 6 const int MAXN = 100017; 7 LL c[MAXN], dp[MAXN]; 8 void init() 9 { 10 memset(dp,0,sizeof(dp)); 11 memset(c,0,sizeof(c)); 12 } 13 int main() 14 { 15 LL n; 16 LL tt; 17 int i, j; 18 while(~scanf("%I64d",&n)) 19 { 20 init(); 21 int maxx = 0; 22 for(i = 1; i <= n; i++) 23 { 24 scanf("%I64d",&tt); 25 if(tt > maxx) 26 maxx = tt; 27 c[tt]++; 28 } 29 dp[0] = 0, dp[1] = c[1]; 30 for(i = 2; i <= maxx; i++) 31 { 32 dp[i] = max(dp[i-1],dp[i-2]+c[i]*i); 33 } 34 printf("%I64d\n",dp[maxx]); 35 } 36 return 0; 37 }
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原文地址：http://www.cnblogs.com/cyd308/p/4650298.html
