# Codeforces - 1321B - Journey Planning(思维)

??思路很简单，因为所得数列满足相邻的两个数的下标之差等于数值之差。所以只要让每个输入的数减去对应的下标所得到的下标指向的数加上这个数即可，如果两个数的下标之差等于数值之差，那么

``````//https://www.cnblogs.com/shuitiangong/
#include<set>
#include<map>
#include<list>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl ‘\n‘
#define rtl rt<<1
#define rtr rt<<1|1
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define zero(a) memset(a, 0, sizeof(a))
#define INF(a) memset(a, 0x3f, sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define _test printf("==================================================\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> P2;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD =  998244353;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6+10;
ll arr[maxn];
int main(void) {
IOS; int n;
const int top = 2e5+10;
while(cin>>n) {
ll ans = 0, num;
for (int i = 1; i<=n; ++i) {
cin >> num;
arr[top+num-i] += num;
ans = max(ans, arr[top+num-i]);
}
cout << ans << endl;
}
return 0;
}
``````

Codeforces - 1321B - Journey Planning(思维)

(0)
(0)