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Circular Merging CodeChef Solution

Circular Merging CodeChef Solution in C++14

``````#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const ll inf=1e18;

/* debugger */
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define dbg(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
/* end of debugger */

void circular_merging() {
int n;
cin>>n;
vector<ll> a(n);
ll tot=0;
for(int i=0;i<n;i++) {
cin>>a[i];
tot+=a[i];
}
vector<vector<ll>> dp(n,vector<ll>(n,inf));
/*
dp[i][j]: cost of merging from index i to j
i can be > j : cyclic
*/
for(int i=0;i<n;i++) {
dp[i][i]=0;
}

for(int len=2;len<=n;len++) {
for(int i=0;i<n;i++) {
int j=(i+len-1)%n;
// merging from i to j
ll sum=0;
for(int k=0;k<len;k++)
sum+=a[(i+k)%n];
for(int k=0;k<len-1;k++) {
int mid=(i+k)%n;
// i to mid, mid+1 to j
dp[i][j]=min(dp[i][j],dp[i][mid]+dp[(mid+1)%n][j]+sum);
}
}
}

ll ans=inf;
for(int len=1;len<n;len++) {
for(int i=0;i<n;i++) {
int j=(i+len-1)%n;
ans=min(ans, dp[i][j]+dp[(j+1)%n][(i-1+n)%n]+tot);
}
}
cout<<ans<<"\n";
}

int main() {
int t;
cin>>t;
while(t--) {
circular_merging();
}
}``````

Circular Merging CodeChef Solution in PYTH 3

``````def toneMerge(tone, n):
if n==2: return sum(tone)  # 长度为2时，就是这两堆石头的和

n = n*2
tone = tone + tone

dp = [[float('inf')]*(n+1) for _ in range(n+1)]  # 初始化dp
sk = [[None]*(n+1) for _ in range(n+1)]  # 初始化sk

for i in range(n+1): dp[i][i], sk[i][i] = 0, i
sm = [tone[0]]  # 记录 列表从0 到 n 的累加和
for i in range(1, n): sm.append(sm[i-1]+tone[i])  # 累加求和

for gap in range(1, n//2):  # i 到 j 的间隔
for i in range(n-gap):
j = i + gap  # 确定本次  i--j
tmp = sm[j] - [0, sm[i-1]][i > 0]  # 获取第 i 到第 j 堆石子的总数量
i1, j1 = sk[i][j-1], sk[i+1][j]+1  # 新的区间
for k in range(i1, j1):  # 选择i 到 j 的最优解
if dp[i][j] > dp[i][k] + dp[k+1][j] + tmp:
dp[i][j], sk[i][j] = dp[i][k] + dp[k+1][j] + tmp, k  # 更新dp， 并记录最有位置k
return min([dp[i][i+n//2-1] for i in range(n//2)])
for _ in range(int(input())):
n = int(input())
lst = list(map(int,input().split()))
print(toneMerge(lst,n))``````

Circular Merging CodeChef Solution in C

``````#include <stdio.h>

#define MAX 100000000000000000

long long int a[1000];
long long int dp[800][800];

long long int min(long long int a, long long int b)
{
return (a>b)?b:a;
}

long long int cirmerge( int i,int j)
{
if(i==j)
return 0;
if(dp[i][j]!= -1)
return dp[i][j];
dp[i][j]= MAX;
for(int k=i;k<=j;k++)
return dp[i][j];
}

int main(void) {
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%lld",&n);
long long int ans=MAX;
for(int i=0;i<n;i++)
{
scanf("%lld",&a[i]);
a[n+i]=a[i];
}
for(int i=1;i<2*n;i++)
for(int i=0;i<2*n;i++)
for(int j=0;j<2*n;j++)
dp[i][j]=-1;
for(int k=0;k<n;k++)
{
ans= min(ans, cirmerge(k,n-1+k));
}

printf("%lld\n",ans);

}

return 0;
}``````

Circular Merging CodeChef Solution in JAVA

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static long find(int arr[],int n){
long sv[][]=new long[n][n];
long dp[][]=new long[n][n];
for(int i=0;i<n;i++) sv[i][i]=arr[i];
for(int col=1;col<n;col++){
for(int row=col-1;row>=0;row--){
dp[row][col]=9223372036854775807L;
sv[row][col]=sv[row][row]+sv[row+1][col];
for(int k=row;k<col;k++) dp[row][col]=Math.min(dp[row][col],dp[row][k]+dp[k+1][col]+sv[row][col]);
}
}
for(int col=0;col<n-1;col++){
for(int row=n-1;row>col;row--){
dp[row][col]=9223372036854775807L;
if(col+1==row) sv[row][col]=sv[0][n-1];
else sv[row][col]=sv[0][n-1] - sv[col+1][row-1];
for(int k=row;k<col+n;k++) dp[row][col]=Math.min(dp[row][col],dp[row][k%n]+dp[(k+1)%n][col]+sv[row][col]);
}
}
long res=dp[0][n-1];
for(int col=1;col<n;col++) res=Math.min(res,dp[col][col-1]);

return res;
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner s=new Scanner(System.in);
int t=s.nextInt();
for(int i=0;i<t;i++){
int n=s.nextInt();
int arr[]=new int[n];
for(int j=0;j<n;j++) arr[j]=s.nextInt();
System.out.println(find(arr,n));
}
}
}``````

Circular Merging CodeChef Solution in PYPY 3

``````def merge(A,p,st,fin):
if st==fin:
p[st][fin]=0
return 0
elif p[st][fin]!=-1:
return p[st][fin]
elif st+1==fin:
p[st][fin]=A[st]+A[fin]
return p[st][fin]

sum1=0
for i in range(st,fin+1):
sum1+=A[i]

min1=10000000000000000
for i in range(st,fin):
tmp=sum1+merge(A,p,st,i)+merge(A,p,i+1,fin)
if min1>tmp:
min1=tmp
p[st][fin]=min1
return min1

t=int(input())
for u in range(t):
n=int(input())
A=[int(x) for x in input().split()]
max=-1
for i in range(n):
if A[i]>max:
max=A[i]
idx=i
A=A[idx:]+A[:idx]
p=[[-1]*n for i in range(n)]
print(merge(A,p,0,n-1))``````

Circular Merging CodeChef Solution in C#

``````using System;

namespace CodeChef19._7
{
// Prepared as a solution to CirMerge
// David Sturge (David_S)  13 July 2019.
// *** use method in editorial, for practice
// https://discuss.codechef.com/t/cirmerge-editorial/29723

class CirMerge
{
static void Main()
{
int space_i;
int t = StringToInt(Console.ReadLine(), 0, out space_i);
while (t-- > 0) {
int n = StringToInt(Console.ReadLine(), 0, out space_i);

long[,] dp = new long[n, n];
long[,] sums = new long[n, n];
int n1 = n - 1;
for (int jj = 0; jj < n; ++jj) {
for (int i = 0; i < n; ++i) {
int j = i + jj;
if (j > n1)
j -= n;
long s;
switch (jj) {
case 0:
dp[i, i] = 0;
sums[i, i] = a[i];
break;
case 1:
s = sums[i, i] + sums[j, j];
sums[i, j] = s;
dp[i, j] = dp[i, i] + dp[j, j] + s;
break;
default:
int k = (i < n1) ?  i + 1 : 0;
s = sums[i, i] + sums[k, j];
sums[i, j] = s;
dp[i, j] = dp[i, i] + dp[k, j];
for (int kk = 1; kk < jj; ++kk) {
int kb = k;
if (++k > n1)
k = 0;
long dpn = dp[i, kb] + dp[k, j];
if (dpn < dp[i, j])
dp[i, j] = dpn;
}
dp[i, j] += s;
break;
}
}
}

long result = dp[0, n1];
for (int i = 1; i < n; ++i) {
if (dp[i, i - 1] < result)
result = dp[i, i - 1];
}

Console.WriteLine(result);
}
}

private static long[] StringToLongArray(int num, String s)
// Decode num integers from line of input
// Store them as 'long' because it will be useful later.
{
long[] output = new long[num];
int start_i = 0;
int len = s.Length;
while (num_read < num && start_i < len) {
int next_i;
output[num_read++] = StringToInt(s, start_i, out next_i);
start_i = next_i + 1;
}
return output;
}

private static int StringToInt(String s, int index, out int space_i)
// Convert string from index up to next space or end to integer.
// Return space_i = index of next space encountered, or off end of string.
{
const char c_zero = '0';
int n = s[index] - c_zero;
int len = s.Length;
while (++index < len) {
if (s[index] < c_zero)
break;
n = n * 10 + ( s[index] - c_zero );
}
space_i = index;
return n;
}
}
}``````

Circular Merging CodeChef Solution in GO

``````package main

import (
"bufio"
"fmt"
"os"
)

func readInt(bytes []byte, from int, val *int) int {
i := from
sign := 1
if bytes[i] == '-' {
sign = -1
i++
}
tmp := 0
for i < len(bytes) && bytes[i] != ' ' {
tmp = tmp*10 + int(bytes[i]-'0')
i++
}
*val = tmp * sign
return i
}

func readNum(scanner *bufio.Scanner) (a int) {
scanner.Scan()
return
}

func readTwoNums(scanner *bufio.Scanner) (a int, b int) {
a, b = res[0], res[1]
return
}

func readNNums(scanner *bufio.Scanner, n int) []int {
res := make([]int, n)
x := 0
scanner.Scan()
for i := 0; i < n; i++ {
for x < len(scanner.Bytes()) && scanner.Bytes()[x] == ' ' {
x++
}
}
return res
}

func fillNNums(scanner *bufio.Scanner, n int, res []int) {
x := 0
scanner.Scan()
for i := 0; i < n; i++ {
for x < len(scanner.Bytes()) && scanner.Bytes()[x] == ' ' {
x++
}
}
}

func readUint64(bytes []byte, from int, val *uint64) int {
i := from

var tmp uint64
for i < len(bytes) && bytes[i] != ' ' {
tmp = tmp*10 + uint64(bytes[i]-'0')
i++
}
*val = tmp

return i
}

func readInt64(bytes []byte, from int, val *int64) int {
i := from
var tmp int64
for i < len(bytes) && bytes[i] != ' ' {
tmp = tmp*10 + int64(bytes[i]-'0')
i++
}
*val = tmp
return i
}

func readNInt64Nums(scanner *bufio.Scanner, n int) []int64 {
res := make([]int64, n)
x := -1
scanner.Scan()
for i := 0; i < n; i++ {
}
return res
}

func main() {
scanner := bufio.NewScanner(os.Stdin)

for tc > 0 {
tc--
fmt.Println(solve(n, A))
}
}

const INF = 1 << 50

func solve(n int, A []int) int64 {
if n < 2 {
return 0
}

if n == 2 {
return int64(A[0] + A[1])
}

B := make([]int, 2*n)
copy(B, A)
copy(B[n:], A)

N := 2 * n
dp := make([][]int64, N)
sum := make([][]int64, N)
for i := 0; i < N; i++ {
dp[i] = make([]int64, N)
sum[i] = make([]int64, N)
for j := 0; j < N; j++ {
dp[i][j] = INF
}
dp[i][i] = int64(A[i%n])
sum[i][i] = int64(A[i%n])
}

for j := 1; j < N; j++ {
for i := j - 1; i >= 0 && i > j-n; i-- {
sum[i][j] = int64(A[i%n]) + sum[i+1][j]
for k := i; k < j; k++ {
dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j])
}
dp[i][j] += sum[i][j]
}
}

var best int64 = INF

for i := 0; i < n; i++ {
for j := i; j < i+n; j++ {
a := dp[i][j]
b := dp[j+1][i-1+n]
best = min(best, a+b)
}
}

return best
}

func min(a, b int64) int64 {
if a <= b {
return a
}
return b
}``````
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