1 条题解
-
0
C :
#include <stdio.h> #include<string.h> #include<math.h> #define maxn 100 int main() { int c, n, i, j; double v = 0, w = 0, tmp = 0; double y = 0, ar[maxn]; memset(ar, 0, sizeof(ar)); scanf("%d", &c); while (c) { i = 0; scanf("%d %lf %lf", &n, &v, &w); while (i<n) { scanf("%lf", &ar[i]); i++; } for (i = 0; i<n; i++) { for (j = 0; j<n; j++) { if (ar[i]<ar[j]) { int t = ar[j]; ar[j] = ar[i]; ar[i] = t; } } } j = 0; for (i = 0; i<n;i++) { tmp = ar[i] * v; if ((y + tmp) / (double)(v*(i + 1)) <= w) { y = y + tmp; j++; } else { break; } } if (j==0) { y = 0; } else { y = y / (v*j); int p = y; if (y-(double)p>=0.5) { y += 0.5; } } printf("%d %.2f\n", (int)(v*j),y / 100.0); y = 0.0; c--; } return 0; }
C++ :
#include<stdio.h> #include<stdlib.h> int cmp(const void *a,const void *b){ return *((int*)a)-*((int*)b); } int main(){ int i,n,v,w,cnt; double now; int as[109],cas; scanf("%d",&cas); while( cas-- ){ scanf("%d%d%d",&n,&v,&w); now=0; cnt=0; for( i=0; i<n; i++ ) scanf("%d",as+i); qsort(as,n,sizeof(as[0]),cmp); for( i=0; i<n; i++ ){ if( now+as[i] > (cnt+1)*w ) break; else now+=as[i],cnt++; } if( cnt ) now=now/cnt/100; printf("%d %.2lf\n",cnt*v,now); } return 0; }
Java :
import java.text.DecimalFormat; import java.util.Arrays; import java.util.Scanner; class Main { private static Scanner scan=new Scanner(System.in); public static void main(String[] args) { // TODO Auto-generated method stub int num=scan.nextInt(); double[][] values=new double[num][2]; for (int i = 0; i < num; i++) { int n=scan.nextInt(); int V=scan.nextInt(); int W=scan.nextInt(); int[] Pi=new int[n]; for (int j = 0; j < n; j++) { Pi[j]=scan.nextInt(); } Arrays.sort(Pi); if(Pi[0]>W) { values[i][0]=0; values[i][1]=0.00; } else if(n==1) { values[i][0]=V; values[i][1]=(double)Pi[0]/100; } else { double result=0.00; int count=0; for (int j = 0; j < n; j++) { if((result + Pi[j]) <= (W * (j + 1))) { result+=Pi[j]; count++; } else { break; } } values[i][0]=V*count; values[i][1]=result/(count*100); } } DecimalFormat df=new DecimalFormat("0.00"); for (int i = 0; i < num; i++) { String s=df.format(values[i][1]); System.out.println((int)values[i][0]+" "+s); } } }
- 1
信息
- ID
- 681
- 时间
- 1000ms
- 内存
- 32MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者