数学考试1

// Problem: 数学考试
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1022
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-03-15 10:24:24
// 
// Powered by CP Editor (https://cpeditor.org)

#include
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair pii;
typedef pair pll;
typedef vector vi;   
typedef vector Vll;               
typedef vector > vpii;
typedef vector > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
void solve () {
	int n, k;
    scanf("%d%d", & n, & k);
    vector a(n + 3, 0),b (n + 3, 0), c(n + 3, 0);
    for (int i = 1; i <= n; i ++)
    {
        scanf("%lld", & a[i]);
        b[i] = a[i];
        c[i] = a[i];
    }
    for (int i = 1; i <= n; i ++)b[i] += b[i - 1];
    for (int i = n; i >= 1; i --) c[i] += c[i + 1];
    
    ll suf = LONG_LONG_MIN, ans = LONG_LONG_MIN;
    for (int i = n - k; i >= k; i --) {
         suf = max (suf, c[i + 1] - c[i + k + 1]);
         ans = max (ans, b[i] - b[i - k] + suf);
    }
    
    print(ans);
    puts("");
}
signed main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

题意:给你一个序列，要求求长度为n，选两个不相交的长度为k的区间的最大和

问题结构:固定区间大小

题解:从后往前遍历，依次维护最大的后缀和，然后对于每个i选择去max
我在想用什么数据结构，实际上这题不需要数据结构