HDU 5355 Cake, hdu5355cake

HDU 5355 Cake, hdu5355cake

HDU 5355 Cake

Updated code:

I thought a lot of ideas when I re-wrote this question today.

Finally, I finally came up with a thought of perfection, but the result timed out.

I feel like I am not saving myself.

Finally, we added the memory-based search and the AC

Okay, let's talk about it first. I don't know if everyone is paying attention to m <= 10.

We can write most of the data in pairs, for example, n = 27 m = 6.

1st copies 27 16

2nd copies 26 17

3rd copies 25 18

4th copies 24 19

5th copies 23 20

6th copies 22 21

1 ~ 15. Search for 6 equal parts to all users.

In this way, we try to add as many even numbers as possible.

Ensure that the remaining number is greater than or equal to 2 * m and less than 4 * m

So we only need to search for the remaining number smaller than 4 m (less than 40 ).

So the range of n is 1 ~ 40, m range 1 ~ 10. record these results in this way (to prevent such data from repeated occurrence)

In this case, if we have data after 15 or 6 calculations

N = 27 m = 6

N = 39 m = 6

N = 15 + (arbitrary 12) m = 6

We do not need to search any more.

In this way, the problem is solved.

# Include <algorithm> # include <iostream> # include <cstring> # include <cstdio> # include <vector> # define maxn 100005 # define ll _ int64 # define cnm c [nu] [m] using namespace std; int v [15] [maxn]; ll m, n; // fun Function Definition divides 1-num evenly to l-r individual int a [15] [50]; int si [15]; bool B [50]; int dp [50] [10] = {0 }; int c [50] [10] [15] [50] = {0}; ll ave, nu; bool dfs (int sum, int g, int s) {if (sum = ave) {g ++; sum = 0; s = 1 ;}if (g = m-1) {for (int I = 1; I <= nu; I ++) if (! B [I]) a [g] [++ a [g] [0] = I; return true;} for (int I = s; I <= nu; I ++) {if (sum + I> ave) return false; if (! B [I]) {B [I] = true; a [g] [++ a [g] [0] = I; if (dfs (sum + I, g, I + 1) return true; B [I] = false; a [g] [0] --;} return false;} bool fun (ll num) {if (num> = 4 m-1) {for (int I = 0, j = 0; I <m; I ++, j ++) {v [I] [si [I] ++] = num-j; v [I] [si [I] ++] = num-2 * m + 1 + j ;} return fun (num-2 * m);} else {nu = num; ave = (num + 1) * num/2/m; int B; if (dp [num] [m] = 0) {if (dfs (0, 0, 1) {dp [num] [m] = 1; for (int I = 0; I <m; I ++) for (int j = 0; j <= a [I] [0]; j ++) cnm [I] [j] = a [I] [j];} else dp [num] [m] =-1 ;} if (dp [nu] [m] = 1) return true; return false ;}} void out (int n) {printf ("YES \ n "); for (int I = 0; I <n; I ++) {printf ("% d", si [I] + cnm [I] [0]); for (int j = 0; j <si [I]; j ++) printf ("% d", v [I] [j]); for (int j = 0; j <cnm [I] [0]; j ++) printf ("% d", cnm [I] [j + 1]); printf ("\ n") ;}} void Init () {memset (si, 0, sizeof (si); memset (a, 0, sizeof ()); memset (B, 0, sizeof (B);} int main () {int T; scanf ("% d", & T); while (T --) {scanf ("% I64d % I64d", & n, & m); ll su = n * (n + 1)/2; ll av = su/m; if (su % m | av <n) printf ("NO \ n"); else {Init (); bool OK = fun (n); if (OK) out (m); else printf ("NO \ n") ;}} return 0 ;}

 

/*

This code has limitations. After data is updated

To be updated ......

*/

The first thought of this question seems to be brute force, and the result is actually over again.

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#define maxn 100005#define ll __int64using namespace std;int a[maxn];vector <int > v[maxn];int main(){    int T;    scanf("%d",&T);    while(T--){        ll n,m;        scanf("%I64d%I64d",&n,&m);        ll sum =(n+1)*n/2;        if(sum%m==0){            ll ave=sum/m;            if(ave<n)                printf("NO\n");            else{                memset(v,0,sizeof(v));                memset(a,0,sizeof(a));                int t=ave,A=0;                int flag=0;                for(int i=n;!flag&&i>=1;i--){                    if(a[i]==0){                        t-=i;                        v[A].push_back(i);                        a[i]=1;                    }                    if(i-1>t){                        for(int j=i-1;t&&j>=1;j--){                            if(t>=j&&a[j]==0){                                v[A].push_back(j);                                t-=j;                                a[j]=1;                            }                        }                    }                    if(t==0){                        t=ave;                        A++;                    }                }                if(A==m){                    printf("YES\n");                    for(int i=0;i<m;i++){                        printf("%d",v[i].size());                        for(int j=0;j<v[i].size();j++)                            printf(" %d",v[i][j]);                        printf("\n");                    }                }                else                    printf("NO\n");            }        }        else            printf("NO\n");    }    return 0;}

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