CodeForces - 1469B Red and Blue (前缀和)

简介: CodeForces - 1469B Red and Blue (前缀和)

CodeForces - 1469B Red and Blue (前缀和

原题链接

思路:

想复杂了。

要知道的是合并后的前缀和等于两个序列的前缀和之和。

然后两个都取最大值就好啦。

代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;typedef unsigned long long ull;
typedef pair<ll,ll>PLL;typedef pair<int,int>PII;typedef pair<double,double>PDD;
#define I_int ll
inline ll read(){ll 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*10+ch-'0';ch=getchar();}return x*f;}
#define read read()
ll ksm(ll a,ll b,ll p){ll res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
const int maxn=1e6+7,inf=0x3f3f3f3f;
int a[maxn],b[maxn],n,m;
void solve(){
  int sum=0,res=0,tmp=0;
  n=read;
  for(int i=1;i<=n;i++) 
    a[i]=read,tmp+=a[i],sum=max(sum,tmp);
  m=read;
  res+=sum;
  tmp=0,sum=0;
  for(int i=1;i<=m;i++) 
    b[i]=read,tmp+=b[i],sum=max(sum,tmp);
  res+=sum;
  cout<<res<<endl;
}
int main(){
  int T=read;
  while(T--) solve(); 
  return 0;
}


目录
相关文章
|
7月前
|
Java
hdu-1312-Red and Black
hdu-1312-Red and Black
35 0
HDU - 1312 Red and Black(DFS)
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles. Write a program to count the number of black
90 0
|
C++
【PAT甲级 - C++题解】1135 Is It A Red-Black Tree
【PAT甲级 - C++题解】1135 Is It A Red-Black Tree
105 0
|
机器学习/深度学习 Windows
Codeforces Round #748 (Div. 3) F - Red-Black Number (记忆化搜索)
Codeforces Round #748 (Div. 3) F - Red-Black Number (记忆化搜索)
104 0
hdu 1312 Red and Black(BFS)
hdu 1312 Red and Black(BFS)
147 0
hdu 1312 Red and Black
一个人从@点出发,求他所能到达的'.'的数目,'#'不可走,@本身算1个点。 思路:搜索入门题。
155 0
HDOJ 1312 (POJ 1979) Red and Black
HDOJ 1312 (POJ 1979) Red and Black
116 0
【1069】The Black Hole of Numbers (20 分)
【1069】The Black Hole of Numbers (20 分) 【1069】The Black Hole of Numbers (20 分)
107 0

热门文章

最新文章