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zju 3175 Number of Containers

DreamSky — Thu, 07 May 2009 04:07:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216

//1858220 2009-05-07 11:33:22 Accepted  3175 C++ 70 184 aslys
// x*y = n ……
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int i;
        long long n,s,sum=0;
        scanf("%lld",&n);
        s = sqrt((double)n);
        for(i = 1;i<=s;i++)
            sum += n/i;
        sum *= 2;
        sum -= s*s;
        sum -= n;
        printf("%lld\n",sum);
    }
    return 0;
}

DreamSky 2009-05-07 12:07 发表评论

DreamSky — Tue, 05 May 2009 04:53:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1267

#include<iostream>
using namespace std;
int main()
{
    int i,j,len=1,r=0,temp=0;
    int a[201][400]={0};
    a[1][0]=1;
    for(i=2;i<=200;i++)//�?……200的阶�?/span>
    {
        for(j=0;j<len;j++)
            a[i][j] = a[i-1][j] * i;
        for(j=0,r=0;j<len;j++)
        {
            temp = a[i][j] + r;
            a[i][j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            a[i][len++] = r%10;
            r /= 10;
        }
    }
    int m,n;
    while(cin>>m>>n)
    {
        r = 0;
        len = 400;//�Ҏ��300��错�?/span>
        int b[500]={0};
        for(j=0;j<len;j++)
        {
            temp = a[m+n][j]*(m+1-n) + r;
            b[j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            b[len++] = r%10;
            r /= 10;
        }

        for(j=len-1,r=0;j>=0;j--)
        {            //除法从高位到低位
            temp=r*10+b[j];
            b[j]=temp/(m+1);
            r=temp%(m+1);
        }
        while(!b[len])        //处理高位的零�?/span>
            len--;

        for(i=m;i>1;i--)//除m!
        {
            for(j=len,r=0;j>=0;j--)
                {            //除法从高位到低位
                temp=r*10+b[j];
                b[j]=temp/i;
                r=temp%i;
                }
            while(!b[len])        //处理高位的零�?/span>
            len--;
        }

        for(i=n;i>1;i--)//除n!
        {
            for(j=len,r=0;j>=0;j--)
                {            //除法从高位到低位
                temp=r*10+b[j];
                b[j]=temp/i;
                r=temp%i;
                }
            while(!b[len])        //处理高位的零�?/span>
            len--;
        }

        for(j=len;j>=0;j--)
            cout<<b[j];
        cout<<endl;
    }
    return 0;
}

DreamSky 2009-05-05 12:53 发表评论

hdu 1023 Train Problem II

DreamSky — Tue, 05 May 2009 03:02:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1023

//C(0) = 1 ; (n+2)*C(n+1) = (4n+2)*C(n);
#include<iostream>
using namespace std;
int a[101][101]={0};
int main()
{
    int n,i,j,len,r,temp,t;
    int b[101];
    a[1][0] = 1;
    len = 1;
    b[1] = 1;
    for(i=2;i<=100;i++)
    {
        t = i-1;
        for(j=0;j<len;j++) //乘法
            a[i][j] = a[i-1][j]*(4*t+2);
        for(r=j=0;j<len;j++)  //处理�怹��l�果
        {
            temp = a[i][j] + r;
            a[i][j] = temp % 10;
            r = temp / 10;
        }
        while(r) //�q�位处理
        {
            a[i][len++] = r % 10;
            r /= 10;
        }

        for(j=len-1,r=0;j>=0;j--) //除法
        {
            temp = r*10 + a[i][j];
            a[i][j] = temp/(t+2);
            r = temp%(t+2);
        }
        while(!a[i][len-1]) //高位零处�?/span>
            len --;
        b[i] = len;
    }
    while(cin>>n)
    {
        for(j=b[n]-1;j>=0;j--)
            printf("%d",a[n][j]);
        printf("\n");
    }
    return 0;
}
/*
1: 1
2: 2
3: 5
4: 14
5: 42
6: 132
7: 429
8: 1430
9: 4862
10: 16796
11: 58786
12: 208012
13: 742900
14: 2674440
15: 9694845
16: 35357670
17: 129644790
18: 477638700
19: 1767263190
20: 6564120420
21: 24466267020
22: 91482563640
23: 343059613650
24: 1289904147324
25: 4861946401452
26: 18367353072152
27: 69533550916004
28: 263747951750360
29: 1002242216651368
30: 3814986502092304
31: 14544636039226909
32: 55534064877048198
33: 212336130412243110
34: 812944042149730764
35: 3116285494907301262
36: 11959798385860453492
37: 45950804324621742364
38: 176733862787006701400
39: 680425371729975800390
40: 2622127042276492108820
41: 10113918591637898134020
42: 39044429911904443959240
43: 150853479205085351660700
44: 583300119592996693088040
45: 2257117854077248073253720
46: 8740328711533173390046320
47: 33868773757191046886429490
48: 131327898242169365477991900
49: 509552245179617138054608572
50: 1978261657756160653623774456
51: 7684785670514316385230816156
52: 29869166945772625950142417512
53: 116157871455782434250553845880
54: 451959718027953471447609509424
55: 1759414616608818870992479875972
56: 6852456927844873497549658464312
57: 26700952856774851904245220912664
58: 104088460289122304033498318812080
59: 405944995127576985730643443367112
60: 1583850964596120042686772779038896
61: 6182127958584855650487080847216336
62: 24139737743045626825711458546273312
63: 94295850558771979787935384946380125
64: 368479169875816659479009042713546950
65: 1440418573150919668872489894243865350
66: 5632681584560312734993915705849145100
67: 22033725021956517463358552614056949950
68: 86218923998960285726185640663701108500
69: 337485502510215975556783793455058624700
70: 1321422108420282270489942177190229544600
71: 5175569924646105559418940193995065716350
72: 20276890389709399862928998568254641025700
73: 79463489365077377841208237632349268884500
74: 311496878311103321137536291518809134027240
75: 1221395654430378811828760722007962130791020
76: 4790408930363303911328386208394864461024520
77: 18793142726809884575211361279087545193250040
78: 73745243611532458459690151854647329239335600
79: 289450081175264899454283846029490767264392230
80: 1136359577947336271931632877004667456667613940
81: 4462290049988320482463241297506133183499654740
82: 17526585015616776834735140517915655636396234280
83: 68854441132780194707888052034668647142985206100
84: 270557451039395118028642463289168566420671280440
85: 1063353702922273835973036658043476458723103404520
86: 4180080073556524734514695828170907458428751314320
87: 16435314834665426797069144960762886143367590394940
88: 64633260585762914370496637486146181462681535261000
89: 254224158304000796523953440778841647086547372026600
90: 1000134600800354781929399250536541864362461089950800
91: 3935312233584004685417853572763349509774031680023800
92: 15487357822491889407128326963778343232013931127835600
93: 60960876535340415751462563580829648891969728907438000
94: 239993345518077005168915776623476723006280827488229600
95: 944973797977428207852605870454939596837230758234904050
96: 3721443204405954385563870541379246659709506697378694300
97: 14657929356129575437016877846657032761712954950899755100
98: 57743358069601357782187700608042856334020731624756611000
99: 227508830794229349661819540395688853956041682601541047340
100: 896519947090131496687170070074100632420837521538745909320
*/

DreamSky 2009-05-05 11:02 发表评论

hdu 2669 Romantic

DreamSky — Mon, 04 May 2009 11:04:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=2669

//�Ƨ几里�d��法�Q�Euclid�Q?/span>
#include <iostream>
using namespace std;
int exgcd(int a, int b, int &x, int &y)//辗�{盔R��法求最大公�U�数GCD
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int r = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return r;
}
int main()
{
    int m,n,a,b,r;
    while(scanf("%d%d",&a,&b)+1)
    {
        r=exgcd(a,b,m,n);
        if(r==1)//最大公�U�数�?�?/span>
        {
            while(m<0)
            {
                m+=b;
                n-=a;
            }
            printf("%d %d\n",m,n);
        }
        else printf("sorry\n");
    }
    return 0;
}

DreamSky 2009-05-04 19:04 发表评论

hdu 1134 Game of Connections

DreamSky — Thu, 30 Apr 2009 07:51:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1134

#include <iostream>
using namespace std;
#define SIZE 100
int main()
{
    int a[101][SIZE]={0}; //数组用来存放�l�果
    a[1][0]=1;            //初始化第一��?nbsp;
    int i,j,r=0,temp=0,len=1;    //len 表示当前最长的有效位数�Q�初始�ؓ1
    for(i=2;i<=100;i++)          //从第二项到第100��，用公式计��?/span>
    {
        for(j=0;j<len;j++)   //---------------乘法部分------------------
        {
            a[i][j]=a[i-1][j]*(4*i-2);    //乘法从低位到高位
        }
        for(j=0;j<len;j++)        //对乘出的�l�果�q�行处理�Q�不包括最高位
        {
            temp=a[i][j]+r;
            a[i][j]=temp%10;
            r=temp/10;
        }
        while(r)            //�Ҏ��高位�q�位处理
        {
            a[i][len]=r%10;
            r/=10;
            len++;
        }
        //-----------------除法部分-----------------
        for(j=len-1,r=0;j>=0;j--)
        {            //除法从高位到低位
            temp=r*10+a[i][j];
            a[i][j]=temp/(i+1);
            r=temp%(i+1);
        }
        while(!a[i][len-1])        //处理高位的零�?/span>
            len--;

    }
    //-------------------------------------------
    int n;
    while(cin>>n&&n!=-1)
    {
        for(i=SIZE-1;!a[n][i];i--);
        for(i;i>=0;i--)
            cout<<a[n][i];
        cout<<endl;
    }
    return 0;
}

DreamSky 2009-04-30 15:51 发表评论

hdu 1133 Buy the Ticket

DreamSky — Thu, 30 Apr 2009 07:36:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1133

//1323739 2009-04-30 15:27:35 Accepted 1133 15MS 580K 1116 B C++ no way
#include<iostream>
using namespace std;
int main()
{
    int i,j,len=1,r=0,temp=0;
    int a[201][400]={0};
    a[1][0]=1;
    for(i=2;i<=200;i++)//�?……200的阶�?/span>
    {
        for(j=0;j<len;j++)
            a[i][j] = a[i-1][j] * i;
        for(j=0,r=0;j<len;j++)
        {
            temp = a[i][j] + r;
            a[i][j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            a[i][len++] = r%10;
            r /= 10;
        }
    }
    int m,n,cas=1;
    while(cin>>m>>n)
    {
        if(m==0 && n==0)
            break;
        cout<<"Test #"<<cas++<<":"<<endl;
        if(n>m)
        {
            cout<<0<<endl;
            continue;
        }
        r = 0;
        len = 400;//�Ҏ��300��错�?/span>
        int b[500]={0};
        for(j=0;j<len;j++)
        {
            temp = a[m+n][j]*(m+1-n) + r;
            b[j] = temp % 10;
            r = temp /10;
        }
        while(r)
        {
            b[len++] = r%10;
            r /= 10;
        }

        for(j=len-1,r=0;j>=0;j--)
        {            //除法从高位到低位
            temp=r*10+b[j];
            b[j]=temp/(m+1);
            r=temp%(m+1);
        }
        while(!b[len])        //处理高位的零�?/span>
            len--;
        for(j=len;j>=0;j--)
            cout<<b[j];
        cout<<endl;
    }
    return 0;
}

DreamSky 2009-04-30 15:36 发表评论

zju 1199 Point of Intersection

DreamSky — Thu, 02 Apr 2009 01:37:00 GMT

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=199

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        double x1,y1,r1,x2,y2,r2,temp,drr,x0,y0;
        cin>>x1>>y1>>r1;
        cin>>x2>>y2>>r2;
        if(r1>r2)//r1始终是小�?/span>
        {
            temp=r1; r1=r2; r2=temp;
            temp=x1; x1=x2; x2=temp;
            temp=y1; y1=y2; y2=temp;
        }
        drr=sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
        if(r2==r1)
        {
            cout<<"Impossible."<<endl;
            continue;
        }
        if(r1+drr<=r2)//内含或者内�?/span>
        {
            cout<<"Impossible."<<endl;
            continue;
        }
        if(x1==0 && x2==0)//两圆心在y轴上
        {
            temp=r1*drr/(r2-r1);
            if(y1>y2)
                printf("0.00 %.2lf\n",y1+temp);
            else
                printf("0.00 %.2lf\n",y1-temp);
            continue;
        }
        if(y1==0 && y2==0)//两圆心在x轴上
        {
            temp=r1*drr/(r2-r1);
            if(x1>x2)
                printf("%.2lf 0.00\n",x1+temp);
            else
                printf("%.2lf 0.00\n",x1-temp);
            continue;
        }

        x0=(r2*x1-r1*x2)/(r2-r1);
        y0=(y1-y2)*r1/(r2-r1)+y1;

        printf("%.2lf %.2lf\n",x0,y0);
    }
    return 0;
}

DreamSky 2009-04-02 09:37 发表评论