#260 (div.1) A.Boredom
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1.题目描述:点击打开链接
2.解题思路:本题属于01背包型的dp问题。为了使得问题变得便与思考,我们可以顺序考虑每一个数字。事先用数组cnt统计输入的数字的个数。接下来,考虑数字i。如果选择删除它的话,由于是按顺序考虑的,因此所有的i-1都将被删去,那么分数为d[i]+i*cnt[i];如果不删除它,那么分数就是d[i-1]。只需要取较大者即可。
3.代码:
#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef long long ll;typedef unsigned long long ull;#define me(s) memset(s,0,sizeof(s))#define For(i,n) for(int i=0;i<(n);i++)#define MAX 1000005int a[MAX];ll d[MAX];int cnt[MAX];int n;int main(){//freopen("t.txt", "r", stdin);while (cin >> n){me(a); me(d); me(cnt);int maxd = 0;For(i, n){cin >> a[i];maxd = max(maxd, a[i]);cnt[a[i]]++;}ll ans = 0;d[1] = cnt[1];for (int i = 1; i <= maxd; i++){if(i>1)d[i] = d[i - 1];if (i - 2 >= 0)d[i] = max(d[i], d[i - 2] + (ll)i*cnt[i]);ans = max(ans, d[i]);}cout << ans << endl;}return 0;}
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