time limit per test
2 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.
The picture corresponds to the first example. Blue boxes illustrate blocks.
Write a program to find such a set of blocks.
Input
The first line contains integer nn (1≤n≤501≤n≤50) — the length of the given array. The second line contains the sequence of elements a[1],a[2],…,a[n]a[1],a[2],…,a[n] (−105≤ai≤105−105≤ai≤105).
Output
In the first line print the integer kk (1≤k≤n1≤k≤n). The following kk lines should contain blocks, one per line. In each line print a pair of indices li,rili,ri (1≤li≤ri≤n1≤li≤ri≤n) — the bounds of the ii-th block. You can print blocks in any order. If there are multiple answers, print any of them.
Examples
input
Copy
7 4 1 2 2 1 5 3
output
Copy
3 7 7 2 3 4 5
input
Copy
11 -5 -4 -3 -2 -1 0 1 2 3 4 5
output
Copy
2 3 4 1 1
input
Copy
4 1 1 1 1
output
Copy
4 4 4 1 1 2 2 3 3
看了别的大佬写的
#include<bits/stdc++.h>
using namespace std ;
int a[55];
int main()
{
int n;
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%d",&a[i]);
map<int,int> mmp,ma;
map<int,int>::iterator it;
//找个数最多的区间和值
for (int i=1;i<=n;i++)//i表示所找区间的右端点
{
int sum = 0;
for (int j=i;j>=1;j--)
{
sum += a[j];
if (mmp[sum]<j)//如果相同的sum而右端点更小则更新，因为这才能保证区间最多
{
mmp[sum] = i;
ma[sum]++;
}
}
}
int ans = -1;
for (it=ma.begin();it!=ma.end();it++)
{
if (it->second>ma[ans])
{
ans = it->first;
}
}
printf("%d\n",ma[ans]);
int r = 0;
//和前面一样，只不过这是遍历区间
for (int i=1;i<=n;i++)
{
int sum = 0;
for (int j=i;j>r;j--)
{
sum += a[j];
if (sum==ans)
{
printf("%d %d\n",j,i);
r = i;
}
}
}
return 0;
}