NOIAC 30 candy-优快云博客

本文探讨了如何通过枚举和二分查找技术，在给定的两个数组中选择元素组合，使得扣除特定权重后的最小愉悦度达到最大。通过维护前缀和并应用二分查找，实现了高效求解。

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思路：

$90pts$:

显然要让最小的愉悦度最大，则维护最大的前缀和，枚举即可。

#include <bits/stdc++.h>
const int INF = 1 << 30;
const int MAXN = 100050;
typedef int intt;
#define int long long
using namespace std;
int n, w, ans = -INF, a[MAXN], b[MAXN], sum1[MAXN], sum2[MAXN];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
intt main() {
    n = read();
    w = read();
    for(int i = 1; i <= n; i++) {
        a[i] = read();
        sum1[i] = sum1[i - 1] + a[i];
    }
    for(int i = 1; i <= n; i++) {
        b[i] = read();
        sum2[i] = sum2[i - 1] + b[i];
    }
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++)
            ans = max(ans, min(sum1[n] - sum1[n - i], sum2[n] - sum2[n - j]) - 1ll * (i + j) * w);
    }
    cout << ans << endl;
    return 0;
}

$100pts$：

我们假设从第一组选，则可以二分那个大于第一组的第二组的最小值，统计答案。

第二组也要这样做一遍，找最大值即可。

#include <bits/stdc++.h>
const int INF = 1 << 30;
const int MAXN = 100050;
typedef int intt;
#define int long long
using namespace std;
int n, w, p1, p2, ans1 = -INF, ans2 = -INF, res, l, r, a[MAXN], b[MAXN], sum1[MAXN], sum2[MAXN];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
bool check1(int x, int y) { return sum2[n] - sum2[n - x] >= y; }
bool check2(int x, int y) { return sum1[n] - sum1[n - x] >= y; }
intt main() {
    n = read();
    w = read();
    for(int i = 1; i <= n; i++) {
        a[i] = read();
        sum1[i] = sum1[i - 1] + a[i];
    }
    for(int i = 1; i <= n; i++) {
        b[i] = read();
        sum2[i] = sum2[i - 1] + b[i];
    }
    for(int i = 1; i <= n; i++) {
        res = sum1[n] - sum1[n - i];
        l = 1, r = n;
        while(l <= r) {
            int mid = (l + r) >> 1;
            if(check1(mid, res)) {
                ans1 = max(ans1, res - (i + mid) * w);
                r = mid - 1;
            }
            else
                l = mid + 1;
        }
    }
    for(int i = 1; i <= n; i++) {
        res = sum2[n] - sum2[n - i];
        l = 1, r = n;
        while(l <= r) {
            int mid = (l + r) >> 1;
            if(check2(mid, res)) {
                ans2 = max(ans2, res - (i + mid) * w);
                r = mid - 1;
            }
            else
                l = mid + 1;
        }
    }
    cout << max(ans1, ans2) << endl;
    return 0;
}