C Candies 前一枚举一个i,求第一行的前i个和第二行从第n个到第i个 代码 cpp include define fi first define se second define pii pair define pdi pair define mp make_pair define pb p
[AtCoder] E - Putting Candies Problem Link If we pick A[i] the 2nd time, it means we have a cycle. Proof: 1st time we pick A[i], the sum before adding A[i] is x; 2nd time we pick A[i], the sum before adding A[i] is y; For this to happen x % N == y % N must...
Choose a pile with the largest number of candies remaining, then eat all candies of that pile. From each pile with one or more candies remaining, eat one candy. The player who eats the last candy on the table, loses the game. Determine which player will win if both players play the ga...
因为奇偶相同的两个数之和一定是偶数,所以YES->sum是偶数。 sum是偶数,那么一定是偶数个奇数加上任意个偶数,偶数个奇数两两合并一下就变成了偶数,n个偶数一定可以通过n-1次替代变成1个数。因此sum是偶数->YES。 代码 代码语言:javascript 代码运行次数:0 #include<cstdio>#include<cstring>#include<algorithm>#...
dp[i][j]表示s前i个字符和t前j个字符所共有的字符的最大个数。最后回溯找答案字符串。 intdp[N][N];voidsolve(){strings,t;cin>>s>>t;intn=s.size(),m=t.size();memset(dp,0,sizeofdp);for(inti=1;i<=n;i++)for(intj=1;j<=m;j++){if(s[i-1]==t[j-1])dp[i][j]=dp[i...
Atcoder beginner contest 087C Candies https://atcoder.jp/contests/abc087/tasks/arc090_a 题意:1个2×n的格子,从左上角出发,每次只能往下或往右走,一直到右下角。求经过的格子中,数值的总和最大值。 思路:直接dp或暴力枚举在第几个格下到第二排。 dp代码: 暴力代码:......
A - Fighting over Candies Solution 看存不存在一个数等于另外两个数的和。 Code #include<algorithm>#include<iostream>usingnamespacestd;typedeflonglongll;inta,b,c;intmain(){cin>>a>>b>>c;if(max({a,b,c})*2==a+b+c)cout<<"Yes\n";elsecout<<"No\n";return0;} ...
There are N boxes arranged in a row from left to right. The i-th box from the leftcontainsAi candies. You will take out the candies from some consecutive boxes and distribute them evenly to M children. Such being the case, find the number of the pairs (l,r) that satisfy the followin...
C - Candies 链接:https://arc090.contest.atcoder.jp/tasks/arc090_a 题意:求矩形左上角到右下角的最大元素和(走动路径只能是向右或者向下) 分析:dfs,也可以是dp 【dfs】代码: #include <bits/stdc++.h> using namespace std; #define mem(a,n) memset(a,n,sizeof(a)) ...
from each dp[st][last], we compute the total distance to reach G, and if it is <= T, then we could update the maximum number of candies that we can take. Is this the intended solution or there is some simpler one. →Reply