Codeforces 453B B. Little Pony and Harmony Chest (dp+ number theory)

Topic Links:

Codeforces 453B

Main topic:

Given a sequence a, the number of a sequence b,b sequence of 22 coprime, ask can cause ∑| a i ? b i | The smallest scenario

Topic Analysis:

• The definition state Dp[i][j] represents the smallest result of the number of previous I to J states, and the J-state represents the prime numbers that have been used.
• Because when the data range of a is not more than 30, so if a number more than 60, then select 1 is better than it, so we can use the number of the quality factor also must not exceed 60, also only 17, so we only need each enumerator, and then the previous state to transfer.
• The specific transfer process is shown in the code.
• In order to get the scheme, we recorded that Pre[i][j] represents the last state where the number of first I reaches the maximum value of the state J, Num[i][j] indicates the number of the number of previous I to the state J to get the maximum value.
  

AC Code:

#include <iostream>#include <cstdio>#include <algorithm>#include <cstdlib>#include <cstring>#define MAX 107using namespace STD;intN,a[max];intdp[107][1<< -];intpre[107][1<< -];intnum[107][1<< -];intprime[]={2,3,5,7, One, -, -, +, at, in, to,Panax Notoginseng, the, A, +, -, -};intstate[ -];voidPrint (intIintu) {if(i = =1)     {printf("%d", Num[i][u]);return; } Print (I-1, Pre[i][u]);printf("%d", Num[i][u]);}intMain () { while( ~scanf("%d", &n)) { for(inti =1; I <= N; i++)scanf("%d", &a[i]);memset(DP,0x3f,sizeof(DP));intINF = dp[0][0]; dp[0][0] =0;memset(State,0,sizeof(state)); for(inti =2; I < -; i + +) for(intj =0; J < -; J + +)if(I%prime[j] = =0) State[i] |= (1&LT;&LT;J);intTotal =1<< -; for(inti =0; I < n; i++) for(intj =0; J < Total; J + +) {if(Dp[i][j] = = INF)Continue; for(intK =1; K < -; k++) {if(STATE[K]&AMP;J)Continue;intx = state[k]|j;if(dp[i+1][X] > Dp[i][j] +ABS(a[i+1]-K) {dp[i+1][X] = Dp[i][j] +ABS(a[i+1]-K); pre[i+1][x] = j; num[i+1][x] = k; }                }            }intAns = Inf,loc =-1; for(inti =0; I < total; i++)if(Ans > Dp[n][i])                {ans = dp[n][i];            loc = i; } print (n, loc);puts(""); }}

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Codeforces 453B B. Little Pony and Harmony Chest (dp+ number theory)