歼灭弱题12

ZOJ 1813 弱
#include <iostream>
#include <iomanip>
using namespace std;
double PI = 3.1415927;
int main()
{
    double d,t,l,s;
    int r,xx=1;
    while(cin >> d >> r >> t)
    {
        if(r==0) break;
        l = d*PI*r/(5280*12);
        s = l*3600/t;
        cout << setiosflags(ios::fixed) << setprecision(2) << "Trip #" << xx++ << ": " << l << " " << s << endl;
    }
    return 0;
}

ZOJ 1402 弱
#include <iostream>
using namespace std;
int main()
{
    int * p;
    int n,i,a,b,fa,fb;
    while(cin >> n)
    {
        if(n==0) break;
        p = new int[n];
        i = 0;a = b = 0;
        while(n--)
        {
            cin >> p[i];
            ++i;
        }
        a = p[0];b = p[i-1];fa = 0;fb = i-1;
        while(fa+1<=fb-1)
        {
            if(a==b)
            {
                a+=p[++fa];
                b+=p[--fb];
            }
            else if(a>b)
            {
                b+=p[--fb];
            }
            else
            {
                a+=p[++fa];
            }
        }
        if(a==b) cout << "Sam stops at position " << fa+1 << " and Ella stops at position " << fb+1 <<"." << endl;
        else cout << "No equal partitioning." << endl;
        delete[] p;
    }
    return 0;
}

ZOJ 1970 相当弱的一个题
#include <iostream>
#include <string>
using namespace std;
int main()
{
    string a,b;
    int la,lb,i,j,t;
    bool f1,f2;
    while(cin >> a >> b)
    {
        f1=true;
        t = -1;
        for(i=0;i<a.size();++i)
        {
            f2=false;
            for(j=t+1;j<b.size();++j)
            {
                if(a[i]==b[j])
                {
                    f2=true;
                    t = j;
                    break;
                }
            }
            if(f2==false)
            {
                f1=false;
                break;
            }
        }
        if(f1==false)
            cout << "No" << endl;
        else
            cout << "Yes" << endl;
    }
    return 0;
}

ZOJ 1904
好短的代码，嘿嘿
不过途中犯了个弱智问题，因为要开3次根号，却在math.h中找不到具体函数，查了半天明白：
原来 pow(double,double)是可以求的，唉，失败
#include <stdio.h>
#include <math.h>
int main()
{
    int N,V;
    while(scanf("%d %d",&N,&V))
    {
        if(N==0&&V==0) break;
        printf("%.3f\n",pow(N*N*N - (6*V)/acos(-1.0),1.0/3));
    }
}