POJ 1877

这一题其实可以写得很简单，只是对于差值的统计比较陌生，连续WA了5，6次才过了，我了个叉。。。

#include <cstdio>using namespace std;const int MAX = 10000;intA[MAX];voidquicksort( int left , int right ){int i = left ;int j = right;int key = A[i];while( i < j ){while( A[j] > key && i < j )--j;if( i < j )A[i++] = A[j];while( A[i] < key && i < j )++i;if( i < j )A[j--] = A[i];}A[i] = key;if( left < i - 1 )quicksort( left , i - 1 );if( i + 1 < right )quicksort( i + 1 , right );}int main(){int row,col,n2;int maxnumber;double sum,mid,high;intk = 1;while( scanf("%d%d",&row,&col) , row != 0 || col != 0 ){n2 = row*col;for( int i = 0 ; i != n2 ; ++i )scanf("%d",&A[i]);scanf("%d",&maxnumber);quicksort( 0 , n2 - 1 );if( k != 1 )printf("/nRegion %d/n",k++);elseprintf("Region %d/n",k++);if( maxnumber == 0 ){printf("Water level is 0.00 meters./n");printf("0.00 percent of the region is under water./n");}else{sum = 0 ;high = 0 ;inti;for( i = 1 ; i < n2 ; ++i ){mid = ( A[i] - A[i-1] ) * i * 100;if( sum + mid >= maxnumber )break;sum += mid;high += A[i] - A[i-1];}high += ( maxnumber - sum )/100.0/i;printf("Water level is %.2lf meters./n",high + A[0]);printf("%.2lf percent of the region is under water./n",i*100.0/n2);}}}