日常刷题2025-2-22

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日常刷题2025-2-22

江

https://ac.nowcoder.com/acm/contest/101963/B

思路：分类谈论

只有两种情况，都考虑一下就行

代码

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
 
using namespace std;
 
void solve(){
    int n; cin>>n;
    vector<int> a(n + 2),b(n + 2);
    for(int i = 1;i<=n;i++) cin>>a[i];
    for(int i = 1;i<=n;i++) cin>>b[i];
    vector<int> p1(n + 2),p2(n + 2),pgx(n + 2,-1e18),sgx(n + 2,-1e18),pmax(n + 2,-1e18),smax(n + 2,-1e18);
    for(int i = 1;i<=n;i++){
        pmax[i] = max(pmax[i-1],b[i]); // 交换分界点上，和左侧时会用到
        p1[i] = p1[i-1] + a[i];
        pgx[i] = b[i] - a[i]; // 交换分界点左右时会用到
        pgx[i] = max(pgx[i],pgx[i-1]);
    }
    for(int i = n;i;i--){
        smax[i] = max(smax[i+1],a[i]); // 交换分界点上，和右侧时会用到
        p2[i] = p2[i+1] + b[i];
        sgx[i] = a[i] - b[i]; // 交换分界点左右时会用到
        sgx[i] = max(sgx[i],sgx[i+1]);
    }
    int ans = -1e18;
    // for(int i = 1;i<=n;i++) cout<<pgx[i]<<" "; cout<<endl;
    for(int i = 1;i<=n;i++){
        ans = max(ans,p1[i]+p2[i]);
        if(i>=2&&i<=n-1) ans = max(ans,p1[i]+p2[i]+pgx[i-1]+sgx[i+1]);
        // cout<<p1[i]+p2[i]+pgx[i-1]+sgx[i+1]<<endl;
        if(i!=n)ans = max(ans,p1[i]+p2[i]-a[i]+smax[i+1]);
        if(i!=1)ans = max(ans,p1[i]+p2[i]-b[i]+pmax[i-1]);
    }
    cout<<ans<<endl;
}
 
signed main(){
    int T; cin>>T; while(T--)
    solve();
    return 0;
}
 

花

https://ac.nowcoder.com/acm/contest/101963/C

思路：数学思维题

第一：此题是计数题，需要考虑在不同长度下，ai 和 aj 的贡献会是多少

第二：在想到第一点还不足以做这道题，因为暴力枚举每个数的组合会超时，所以要考虑枚举优化。此题中的公式在长度相同时是有结合性的，一个一个计算和一次计算一堆是一样的。

代码

#include <bits/stdc++.h>
 
typedef std::pair<long long, long long> pll;
typedef std::pair<int, int> pii;
#define INF 0x3f3f3f3f
#define mod 998244353
using i64 = long long;
const int N = 1e5+5;
 
void solve(){
	int n;
	std::cin >> n;
 
	std::vector<i64> a(n+1);
	for (int i = 1; i <= n; i++){
		std::cin >> a[i];
	}
 
	std::vector <i64> x(n + 1);
    std::vector <i64> y(n + 1);
	std::vector<i64> dpx(n+1), dpy(n+1);
	dpx[1] = 1, dpx[2] = 0;
	dpy[1] = 0, dpy[2] = 1;
	for (int i = 3; i <= n; i++){
		dpx[i] = dpx[i-1] + dpx[i-2];
		dpx[i] %= mod;
		dpy[i] = dpy[i-1] + dpy[i-2];
		dpy[i] %= mod;
	}
 
	for (int i = 1; i <= n; i++) {
        x[n - i + 1] += a[i];
        x[n-i+1] %= mod;
        y[i] += a[i];
        y[i] %= mod;
    }
    for (int i = n - 1; i >= 1; i--)y[i] += y[i + 1], y[i] %= mod;
    for (int i = n - 1; i >= 1; i--)x[i] += x[i + 1], x[i] %= mod;
 
	i64 ans = 0;
	ans += x[1];
	for (int i = 2; i <= n; i++){
		ans += (x[i]*dpx[i]) % mod + (y[i]*dpy[i]) % mod;
		ans %= mod;
	}
 
	std::cout << ans << '\n';
}
 
signed main()
{
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);
	int t = 1, i;
	std::cin >> t;
	for (i = 0; i < t; i++){
		solve();
	}
	return 0;
}