hdu 1081 To The Max
2012-05-18 21:45
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To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4437 Accepted Submission(s): 2099
[align=left]Problem Description[/align]
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle.
In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
[align=left]Input[/align]
The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace
(spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will
be in the range [-127,127].
[align=left]Output[/align]
Output the sum of the maximal sub-rectangle.
[align=left]Sample Input[/align]
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
[align=left]Sample Output[/align]
15
[align=left]Source[/align]
Greater New York 2001
可以根据最大子段,引申出来
#include<cstdio> #include<cstring> #include<algorithm> #define N 50008 using namespace std; int prim ,numprim ; bool isnoprime ; int next; void primm() { memset(isnoprime,0,sizeof(isnoprime)); next=0; int i,j,k; for(i=2;i<N;i++) { if(!isnoprime[i]) {prim[++next]=i;} for(j=i+i;j<N;j++) { isnoprime[j]=1; } } } void DP() { memset(numprim,999999,sizeof(numprim)); numprim[0]=0; numprim[1]=0; int k; for(int i=0;i<N;i++) { for(int j=1;j<=next&&(i+prim[j])<=N;j++) { numprim[i+prim[j]]=min(numprim[i+prim[j]],numprim[i]+1); } for(int j=1;j<=next&&i*prim[j]<=N;j++) numprim[i*prim[j]]=min(numprim[i*prim[j]],numprim[i]+1); } } int main() { int m; scanf("%d",&m); primm(); DP(); int a,b; while(m--) { scanf("%d%d",&a,&b); int sum=0; for(int i=a;i<=b;i++) sum+=numprim[i]; printf("%d\n",sum); } }
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