HDU 5355 Cake (construction + backtracking)

Test instructions

Give a cake, cut into 1~n size n blocks, and ask if you can piece together the M equal without continuing the cut.

Analytical:

First you can find out the sum of these cakes n ( n + 1 ) / 2 , if the sum sum%m! = 0 Then would not be able to be flattened into M-parts, then output "NO".
Next calculate the average avg =sum/m , if the average AVG < N, the cake is impossible to run out of, also output "NO".
The rest of the situation cake is bound to be able to spell "YES", then these cakes can be 2*m as a unit of a group of distribution, everyone take the current group of the largest and smallest, the second and small ....
Until we get the rest. < Span class= "Mrow" id= "mathjax-span-245" > [ 0 4 m Between the cakes is a stop.
At this time of the brute force solution, the problem becomes 1~n number, can spell out, and the same m number, here Direct violence Search DFS.

Summarize:

This problem special judge in the game, so a lot of WA's code is also AC. Later the data is strengthened, and the following code is code that can be enhanced with data.

y code

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#define PB Push_backusing namespace STD;typedef Long LongllConst intN =1e5+Ten; vector<int>ans[ the];intsumv[ the], cake[n];BOOLVis[n];ll N, M;ll sum, AVG, res, dis;voidInit () {memset(SUMV,0,sizeof(SUMV));memset(Vis,false,sizeof(VIS));memset(Cake,0,sizeof(cake)); for(inti =0; I <= m; i++) Ans[i].clear ();}BOOLDfsintCurintSumintPOS) {if(cur = = m +1)return true; for(inti = res; I >= pos; i--) {if(Vis[i])Continue;if(sum + i = = dis)            {Cake[i] = cur; Vis[i] =true;if(Dfs (cur+1,0,1))return true; Vis[i] =false;return false; }Else if(Sum + i < dis)            {Cake[i] = cur; Vis[i] =true;if(Dfs (cur, sum+i, i+1))return true; Vis[i] =false; }     }return false;}intMain () {intTscanf("%d", &t); while(t--) {scanf("%lld%lld", &n, &m); Init ();if(N &1) sum = (n +1) /2NElsesum = N/2* (n +1);if(sum% m! =0) {puts("NO");Continue; } avg = sum/m;if(Avg < N) {puts("NO");Continue; }puts("YES"); res = n% (2* m);if(Res! =0) {res + =2* m;        res = MIN (n, res); }//Divided into 2 * M parts        intA, B; for(inti = n; i > Res; I-= (2* m)) { for(intK =1; K <= m; k++) {a = I-k +1, B = i-(2* m) + K;                ANS[K].PB (a), ANS[K].PB (b);            Sumv[k] + = A, sumv[k] + = b; }} dis = avg-sumv[1]; Dfs1,0,1); for(inti =1; I <= Res;        i++) {ANS[CAKE[I]].PB (i); } for(inti =1; I <= m; i++) {intSize = Ans[i].size ();printf("%d", size); for(intK =0; K < size; k++) {printf("%d", Ans[i][k]); }puts(""); }    }return 0;}

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HDU 5355 Cake (construction + backtracking)