AtCoder Beginner Contest 133 -D — Rain Flows into Dams

https://atcoder.jp/contests/abc133/tasks/abc133_d

思路:每座山为2Xi,每个坝为Ai。已知Ai,求出2Xi。

根据已知的X1,则可分别求出X2-n

代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int a[maxn];
long long ans[maxn];
int main()
{
    std::ios::sync_with_stdio(false);
    int n;
    cin >> n;
    for(int i=1;i<=n;i++)
    {
        cin >> a[i];
    }
    long long sum=0;
    for(int i=1;i<=n-1;i++)
    {
        sum+=a[i];
        if(i&1)
            sum-=2*a[i];
    }
    sum=a[n]-sum;
    sum/=2;
    ans[1]=sum;
    for(int i=2;i<=n;i++)
    {
        ans[i]=a[i-1]-ans[i-1];
    }
    cout << 2*ans[1];
    for(int i=2;i<=n;i++)
        cout << " " << 2*ans[i];
    return 0;
}

原文地址:https://www.cnblogs.com/wjc2021/p/11154246.html

时间: 2024-08-01 03:12:08

AtCoder Beginner Contest 133 -D — Rain Flows into Dams的相关文章

AtCoder Beginner Contest 133 F - Colorful Tree

题意:给出一棵n个点的树,每条边有颜色和边长两个属性,n个询问,每次询问把颜色x的边的边长变为y问u到v的路径长度是多少,询问之间独立. 解法:这道题蛮有意思的.解法参考https://www.cnblogs.com/Tieechal/p/11185912.html这位大佬的,这里说下我的理解. 对于每组询问(x,y,u,v)答案比较显然就是dist(u,v)+(sumlen[x]-sumcnt[x]*y),但是这道题在线不好做我们考虑离线做.但是答案的式子是设计到两个点的,怎么才能离线做呢?这

AtCoder Beginner Contest 103 D(贪心)

AtCoder Beginner Contest 103 D 题目大意:n个点,除第n个点外第i与第i+1个点有一条边,给定m个a[i],b[i],求最少去掉几条边能使所有a[i],b[i]不相连. 按右端点从小到大排序,如果当前选的去掉的边在区间内,那么符合条件,否则ans++,并贪心地把去掉的边指向右端点,因为前面的区间都满足条件了,所以要去掉的边要尽量向右移使其满足更多的区间. 1 #include <iostream> 2 #include <cstdio> 3 #incl

AtCoder Beginner Contest 136

AtCoder Beginner Contest 136 Contest Duration : 2019-08-04(Sun) 20:00 ~ 2019-08-04(Sun) 21:40 Website: AtCoder BC-136 后面几题都挺考思考角度D. C - Build Stairs 题目描述: 有n座山从左到右排列,给定每一座山的高度\(Hi\),现在你可以对每座山进行如下操作至多一次:将这座山的高度降低1. 问是否有可能通过对一些山进行如上操作,使得最后从左至右,山的高度呈不下降

AtCoder Beginner Contest 154 题解

人生第一场 AtCoder,纪念一下 话说年后的 AtCoder 比赛怎么这么少啊(大雾 AtCoder Beginner Contest 154 题解 A - Remaining Balls We have A balls with the string S written on each of them and B balls with the string T written on each of them. From these balls, Takahashi chooses one

AtCoder Beginner Contest 155 简要题解

AtCoder Beginner Contest 155 A:签到失败,WA一次. int main() { int a, b, c; cin >> a >> b >> c; if(a == b && b == c) cout << "No"; else if(a == b || a == c || b == c) cout << "Yes"; else cout << &quo

AtCoder Beginner Contest 152 - F - Tree and Constraints (容斥定理+树上路径的性质)

AtCoder Beginner Contest 152 - F - Tree and Constraints (容斥定理+树上路径的性质) We have a tree with NN vertices numbered 11 to NN. The ii-th edge in this tree connects Vertex aiai and Vertex bibi. Consider painting each of these edges white or black. There ar

【ATcoder】AtCoder Beginner Contest 161 题解

题目链接:AtCoder Beginner Contest 161 原版题解链接:传送门 A - ABC Swap 这题太水,直接模拟即可. 1 #include <iostream> 2 using namespace std; 3 int main() { 4 int a, b, c; 5 cin >> a >> b >> c; 6 swap(a, b); 7 swap(a, c); 8 cout << a << " &

AtCoder Beginner Contest 115 题解

题目链接:https://abc115.contest.atcoder.jp/ A Christmas Eve Eve Eve 题目: Time limit : 2sec / Memory limit : 1024MB Score : 100 points Problem Statement In some other world, today is December D-th. Write a program that prints Christmas if D=25, Christmas E

AtCoder Beginner Contest 106 ABCD

A - Garden Problem Statement There is a farm whose length and width are A yard and B yard, respectively. A farmer, John, made a vertical road and a horizontal road inside the farm from one border to another, as shown below: (The gray part represents