标签：Now Polycarp 题解 CF1512F position tugriks Education day he
题目链接
题目
有一个长度为 \(n\) 的数组 \(a\) 和一个长度为 \(n−1\) 的数组 \(b\)，初始位置为 \(pos=1\)，每一天可以选择得到 \(a_{pos}\)元钱，或者花费 \(b_{pos}\)元钱（钱数不能为负）使得 \(pos\leftarrow pos+1\)
现在希望买一台 \(c\) 元的电脑，最少需要多少天。
Polycarp is wondering about buying a new computer, which costs $ c $ tugriks. To do this, he wants to get a job as a programmer in a big company.
There are $ n $ positions in Polycarp's company, numbered starting from one. An employee in position $ i $ earns $ a[i] $ tugriks every day. The higher the position number, the more tugriks the employee receives. Initially, Polycarp gets a position with the number $ 1 $ and has $ 0 $ tugriks.
Each day Polycarp can do one of two things:
- If Polycarp is in the position of $ x $ , then he can earn $ a[x] $ tugriks.
- If Polycarp is in the position of $ x $ ( $ x < n $ ) and has at least $ b[x] $ tugriks, then he can spend $ b[x] $ tugriks on an online course and move to the position $ x+1 $ .
For example, if $ n=4 $ , $ c=15 $ , $ a=[1, 3, 10, 11] $ , $ b=[1, 2, 7] $ , then Polycarp can act like this:
- On the first day, Polycarp is in the $ 1 $ -st position and earns $ 1 $ tugrik. Now he has $ 1 $ tugrik;
- On the second day, Polycarp is in the $ 1 $ -st position and move to the $ 2 $ -nd position. Now he has $ 0 $ tugriks;
- On the third day, Polycarp is in the $ 2 $ -nd position and earns $ 3 $ tugriks. Now he has $ 3 $ tugriks;
- On the fourth day, Polycarp is in the $ 2 $ -nd position and is transferred to the $ 3 $ -rd position. Now he has $ 1 $ tugriks;
- On the fifth day, Polycarp is in the $ 3 $ -rd position and earns $ 10 $ tugriks. Now he has $ 11 $ tugriks;
- On the sixth day, Polycarp is in the $ 3 $ -rd position and earns $ 10 $ tugriks. Now he has $ 21 $ tugriks;
- Six days later, Polycarp can buy himself a new computer.
Find the minimum number of days after which Polycarp will be able to buy himself a new computer.
思路
从前往后枚举每个位置，设其为整个过程到达的最右的位置。
枚举的过程可以记录当前过程中 \(a_i\) 的最大值，然后当前过程中加的话只加这个词。
到达每个位置后在计算一下当前要达到 \(c\) 的最小值，最后再取个最小值就是答案。
Code
// Problem: CF1512F Education
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF1512F
// Memory Limit: 250 MB
// Time Limit: 2000 ms
#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define N 200010
//#define M
//#define mo
int n, m, i, j, k, T;
int a[N], b[N], d, s, ans, mx, c;
int Z(int x, int y)
{
return x/y+(x%y ? 1 : 0);
}
signed main()
{
// freopen("tiaoshi.in","r",stdin);
// freopen("tiaoshi.out","w",stdout);
T=read();
while(T--)
{
d=-1; s=mx=0; ans=0x7fffffff;
n=read(); c=read();
for(i=1; i<=n; ++i) a[i]=read();
for(i=2; i<=n; ++i) b[i]=read();
for(i=1; i<=n; ++i)
{
if(s<b[i]) d+=Z(b[i]-s, mx), s+=Z(b[i]-s, mx)*mx;
++d; s-=b[i]; mx=max(mx, a[i]);
if(s<c) ans=min(ans, d+Z(c-s, mx));
// printf("%lld %lld %lld\n", s, d, ans);
}
printf("%lld\n", ans);
}
return 0;
}
总结
这道题一开始盲猜是dp，最后发现其实是个很简单的贪心。
核心思想中很重要的一个，就是当前过程如果加就只在当前最大值停留。
感觉和以前的一道什么湖边钓鱼很像。
标签：Now,Polycarp,题解,CF1512F,position,tugriks,Education,day,he 来源： https://www.cnblogs.com/zhangtingxi/p/16365234.html
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