PRoblem Description

提示：多種藥水混合后的濃度=(濃度1*體積1+濃度2*體積2+……)/(體積和)

Input

Output

Example Input

31 100 101002 100 2420 303 100 2420 20 30
Example Output
0 0.00100 0.20300 0.23#include<stdio.h>int main(){    int c,i,j,k;    double n,v,w,pi[101],sum,t1,t2,t;    scanf("%d",&c);    for(i=0;i<c;i++)    {        sum=0;t1=0;t2=0;        scanf("%lf %lf %lf",&n,&v,&w);        for(j=0;j<n;j++)            scanf("%lf",&pi[j]);        for(j=0;j<n;j++)        {            for(k=0;k<n-j-1;k++)            {                if(pi[k]>pi[k+1])                {                    t=pi[k];pi[k]=pi[k+1];pi[k+1]=t;                }            }        }        for(j=0;j<n;j++)        {            t1+=pi[j];            t2+=v;            sum=(t1*v)/t2;            if(j==0&&sum>w)            {                t2=0;                sum=0;                break;            }            if(sum>w)            {                t1=t1-pi[j];                t2=t2-v;                sum=(t1*v)/t2;                break;            }        }        printf("%.0f %.2f/n",t2,sum/100);    }    return 0;}或者用g++提交#include<stdio.h>#include<algorithm>using namespace std;    int pi[110];    int main()    {        int c,n,v,w;        int i,j;        scanf("%d",&c);        while(c--)        {            scanf("%d%d%d",&n,&v,&w);            for(i=0;i<n;i++)                scanf("%d",&pi[i]);            sort(pi,pi+n);            double P=0;//濃度            int V=0; //當前的體積            for(i=0;i<n;i++)            {                if(P*V+pi[i]*v <= w*(V+v))                {                    P=(P*V+pi[i]*v)/(V+v);                    V+=v;                }                else                    break;            }            printf("%d %.2lf/n",V,P/100);        }        return 0;    }