while(ans < n){ if(a[ans] <= t){ t -= a[ans]; ++ ans; }else{ break; } } cout << ans +1<<' '<< t << endl; return0; } D - Max Multiple (abc281 d) 题目大意 给定nn个数aiai,问从中取kk个数出来,其和是dd的倍数,问该和最大是多少。 解题思路 数不大,设dp[i][j][...
那么我们可以进行分治思考,假如该位是第m位了,那么我们取A块与B块在(0,1,,,m-1)中的最小那个; 详细见代码: #include<bits/stdc++.h>#defineint long longintcnt[32][155555];intDFS(intx,intL,intR)//当前位是x{intcur =L;for(inti = L; i <= R; ++i)//该位是0的元素if(!(cnt[x +1...
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AtCoder Beginner Contest 359 (A-G个人题解) 状压DP|单调栈|反悔贪心|树上启发式合并 Stargazer 萌新acmer icpc/ccpc非金牌选手 前言:回归一下abc,这场感觉算是比较edu的一场,简单写个题解 比赛链接: A - Count Takahashi #include <bits/stdc++.h> using namespa…阅读全文 赞同13 4 条...
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Problem A. My Last ABC Problem 只想出了暴力 dp。。。关键是在于考察区间中不同的相邻位置的数量。。 设为x,答案就是这个值的 ceil(x/2),证明可以用鸽巢原理。 Problem B. Arrange Your Balls 反而我觉得比 A 简单,从大到小排序,dfs() 构造即可。
5. [ABC281Ex] AlchemyProblem Link给定X 种1 级宝石,对于一个 i 级宝石可以以如下方式合成:2∼i−1 级的宝石各选 ≤1 个,1 级宝石选任意个不同种的,求能合成多少种 n 级宝石。 数据范围:n≤2×105。设ai 表示能合成的 i 级宝石的数量,容易得到 ai=[zi](1+z)X∏j=2i−1(1+ajz)。
[0]+a[i]);//还要记得防止i和0在同一个集合 dp[i][0][1]=min(dp[i-1][0][1]+b[i-1]+a[i],dp[i-1][1][1]+a[i]); } int ans=inf; for (int i=0;i<=1;i++) { for (int j=0;j<=1;j++) { if(i==j) { ans=min(ans,dp[n][i][j]+b[n]); } else { ...
1449 ABC081B - Shift only C++ 17 (gcc 12.2) AtCoder *200 Aug/19/2024 22:04 1448 A - Entrance Examination C++ 17 (gcc 12.2) AtCoder *100 Aug/19/2024 22:03 1447 D - Pedometer C++ 17 (gcc 12.2) AtCoder *400 Aug/17/2024 19:04 1446 C - Enumerate Sequences C++ 17 (gcc 12.2...
1249 A - ABC String GNU C++17 bitmasks brute force implementation *900 Mar/02/2021 20:51 1248 E - Fib-tree GNU C++17 brute force dfs and similar divide and conquer number theory trees *2400 Mar/02/2021 03:40 1247 D - Genius's Gambit GNU C++17 bitmasks constructive algorithms greedy...