手机移动端网站怎么做的济南seo关键词优化方案

题目链接

技能升级

个人思路

需要给n个技能添加技能点，无论技能点加成如何衰减，每次始终都是选择当前技能加点加成最高的那一项技能，所以最后一次的加点一定也是加在当时技能攻击加成最高的那个。此时，我们去寻找最后一次的加点的攻击力加成的值。
详细思路过程请看Java代码的注释…

参考代码(Java/Cpp)

Java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;public class Main {static int n;static long m;static long[][] arr;// 快速读入对象,此处不用快读,最后几个数据点过不了,拿不足分数static StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));public static int nextInt() throws IOException {st.nextToken();return (int) st.nval;}public static long nextLong() throws IOException {st.nextToken();return (long) st.nval;}public static void main(String[] args) throws IOException {// 技能数量n = nextInt();// 加点次数,根据数据范围得为 longm = nextLong();// arr[i][0] 为第 i 个 技能初次加点的攻击力加成// arr[i][1] 为第 i 个 技能加点的衰减数arr = new long[n][2];for(int i = 0; i < n; ++i) {arr[i][0] = nextLong();arr[i][1] = nextLong();}// 查找最后一次加点时,所增加的攻击力,采用 左闭右闭区间int left = 0, right = 1000000; // a_i的范围while(left <= right) {int mid = (left + right) / 2;// 如果当前情况可加点次数 ≥ 限制次数 m,则 增大最后一次加点数if (check(mid)) {left = mid + 1;} else {right = mid - 1;}}// cnt 计算当前已经加点的次数, sum 计算当前攻击力long cnt = 0, sum = 0;for(int i = 0; i < n; ++i) {if(arr[i][0] < right) continue;long k = (arr[i][0] - right) / arr[i][1] + 1; // 通过等差数列的形式,计算这个技能衰减能够加点的次数cnt += k;sum += (arr[i][0] + (arr[i][0] - (k - 1) * arr[i][1])) * k / 2; // 等差数列求和}sum += (m - cnt) * right; // 可能会出现最后一次加点的值在多个技能里同时出现,并且该数量超过可以加点的限制次数 m,通过该方法减去多加的技能点System.out.println(sum);}static boolean check(long x) {long num = 0;for(int i = 0; i < n; ++i) {if (arr[i][0] < x) continue;num += (arr[i][0] - x) / arr[i][1] + 1; // 等差数列,求ai变成x需要经过几次,并记上当前ai}return num >= m;}
}

Cpp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 3;int n;
ll m, a[N], b[N];int check(int x)
{ll cal = 0;for(int i = 0; i < n; ++i){if(a[i] < x) continue;cal += (a[i] - x) / b[i] + 1;}return cal >= m;
}int main()
{ios::sync_with_stdio(0);cin.tie(0);cin >> n >> m;for (int i = 0; i < n; ++i)cin >> a[i] >> b[i];int left = 0, right = 1e6;while(left <= right){int mid = (left + right) / 2;if(check(mid))left = mid + 1;elseright = mid - 1;}ll cnt = 0, res = 0;for(int i = 0; i < n; ++i){if(a[i] < right) continue;int k = (a[i] - right) / b[i] + 1;cnt += k;res += (a[i] * 2 - (k - 1) * b[i]) * k / 2;}res += (m - cnt) * right;cout << res;
}