Multiplication puzzletime limit: 1000 msmemory limit: 65536 kbthis problem will be judged on PKU. Original ID: 1651
64-bit integer Io format: % LLD Java class name: Main the multiplication puzzle is played with a row of cards, each containing a single positive integer. during the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on right of it. it is not allowed to take out the first and the last card in the row. after the final move, only two cards are left in the row.
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500 + 5000 + 2500 = 8000
If he wocould take the cards in the opposite order, I. e. 50, then 20, then 1, the score wocould be
1*50*20 + 1*20*5 + 10*1*5 = 1000 + 100 + 50 = 1150. Inputthe first line of the input contains the number of cards N (3 <= n <= 100 ). the second line contains N integers in the range from 1 to 100, separated by spaces. outputoutput must contain a single integer-the minimal score. sample Input
610 1 50 50 20 5
Sample output
3650
Sourcenortheastern Europe 2001, far-Eastern subregion problem solving: DP [I] [J] indicates the minimum value after I to J is divided! Why DP [I] [J] = min (DP [I] [J], DP [I] [k] + dp [k] [J] + d [I] * d [J] * d [k]) example 1 2 3 4 5 DP [1] [5] = min (DP [1] [5], DP [1] [3] + dp [3] [5] + d [1] * d [3] * d [5]) DP [I] [J] indicates that I j is left in the I j segment. Like the transfer equation just now, DP [1] [5] is 1 5 after 3?
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define INF 0x3f3f3f3f15 using namespace std;16 int dp[110][110],d[110],n;17 int main(){18 int i,j,k;19 while(~scanf("%d",&n)){20 for(i = 1; i <= n; i++)21 scanf("%d",d+i);22 memset(dp,0,sizeof(dp));23 for(k = 3; k <= n; k++){24 for(i = 1; i+k-1 <= n; i++){25 dp[i][i+k-1] = INF;26 for(j = i+1; j < i+k; j++)27 dp[i][i+k-1] = min(dp[i][i+k-1],dp[i][j]+dp[j][i+k-1]+d[i]*d[j]*d[i+k-1]);28 }29 }30 cout<<dp[1][n]<<endl;31 }32 return 0;33 }
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