Codeforces Round #547 (Div. 3)F2. Same Sum Blocks (Hard)
Codeforces Round #547 (Div. 3)F2. Same Sum Blocks (Hard)
glm233
发布于 2020-09-28 10:39:54
发布于 2020-09-28 10:39:54
F2. Same Sum Blocks (Hard)
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
This problem is given in two editions, which differ exclusively in the constraints on the number nn.
You are given an array of integers a[1],a[2],…,a[n].a[1],a[2],…,a[n]. A block is a sequence of contiguous (consecutive) elements a[l],a[l+1],…,a[r]a[l],a[l+1],…,a[r] (1≤l≤r≤n1≤l≤r≤n). Thus, a block is defined by a pair of indices (l,r)(l,r).
Find a set of blocks (l1,r1),(l2,r2),…,(lk,rk)(l1,r1),(l2,r2),…,(lk,rk) such that:
- They do not intersect (i.e. they are disjoint). Formally, for each pair of blocks (li,ri)(li,ri) and (lj,rj(lj,rj) where i≠ji≠j either ri<ljri<lj or rj<lirj<li.
- For each block the sum of its elements is the same. Formally, a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]=a[l1]+a[l1+1]+⋯+a[r1]=a[l2]+a[l2+1]+⋯+a[r2]= ⋯=⋯= a[lk]+a[lk+1]+⋯+a[rk].a[lk]+a[lk+1]+⋯+a[rk].
- The number of the blocks in the set is maximum. Formally, there does not exist a set of blocks (l′1,r′1),(l′2,r′2),…,(l′k′,r′k′)(l1′,r1′),(l2′,r2′),…,(lk′′,rk′′)satisfying the above two requirements with k′>kk′>k.
题意：在一个给定序列中找到最大数目的不交叉子段和相同
思路：数据范围很小1500，可以用o（n^2）复杂度预处理出所有子段和，存入vector中，对于一个特定的sum相同的各种子段
采用贪心策略，[l,r]子段，取有边界尽量小的，因为要数目最多嘛
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#include<unordered_set>
#define rg register ll
#define inf 2147483647
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define ll long long
#define maxn 200005
const double eps = 1e-8;
using namespace std;
inline ll read()
{
char ch = getchar(); ll s = 0, w = 1;
while (ch < 48 || ch>57) { if (ch == '-')w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
return s * w;
}
inline void write(ll x)
{
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
putchar(x % 10 + 48);
}
inline bool cmp(const pair<ll,ll>&a,const pair<ll,ll>&b)
{
return a.second<b.second;
}
int main()
{
ll n=read();
vector<ll>a(n+5);
map<ll,vector<pair<ll,ll>>>p;
for(rg i=1;i<=n;i++)a[i]=read();
for(rg i=1;i<=n;i++)
{
ll sum=0;
for(rg j=i;j<=n;j++)
{
sum+=a[j];
p[sum].push_back(make_pair(i,j));
}
}
ll ans=0;
vector<pair<ll,ll>>res;
for(auto it:p)
{
vector<pair<ll,ll>>now=it.second;
sort(now.begin(),now.end(),cmp);
vector<pair<ll,ll>>temp;
ll cur=0,r=-1;
for(auto itt:now)
{
if(itt.first>r)
{
cur++;
temp.push_back(itt);
r=itt.second;
}
}
if(cur>ans)
{
ans=cur;
res=temp;
}
}
cout<<ans<<endl;
for(auto it:res)
{
cout<<it.first<<" "<<it.second<<endl;
}
return 0;
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原始发表：2019/08/09 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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