2012/8/9 关于今天的比赛

A 大数运算

大数模版写得不熟，然后出了不少错误，准备自己写一个模版。

 

G 拓展欧几里德求逆元

这道题公式没推错，代入参数搞错了，然后一直在调，浪费不少时间。

 

I 求线段交点

很普通的计算题，WA了6次，最后才A掉，计算时没有考虑double比较大小的细节，手写的fabs没过。

 

今天就A了三道最水的题，比较让我失望，I题不应该错这么多次，细节决定成败，这句话放在哪都没错，因为这样WA掉罚时真的很不值。希望明天比赛顺利，然后进入校队。

A：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<iostream>

using namespace std;



const int base = 10000;

const int width = 4;

const int N = 11000;

char buf[11000];

struct bint

{

    int ln, v[N];

    bint(int r = 0){

        for(ln = 0; r > 0; r /= base) v[ln ++] = r % base;

    }

    bint& operator = (const bint& r){

        memcpy(this, &r, (r.ln + 1) * sizeof(int));

        return * this;

    }

};



bint a, b, c;



bool operator < (const bint& a, const bint& b){

    int i;

    if(a.ln != b.ln) return a.ln < b.ln;

    for(i = a.ln - 1; i >= 0 && a.v[i] == b.v[i]; i --);

    return i < 0 ? 0 : a.v[i] < b.v[i];

}



bool operator <= (const bint& a, const bint& b){

    return !(b < a);

}



bint operator + (const bint& a, const bint& b)

{

    bint res;

    int i, cy = 0;

    for(i = 0; i < a.ln || cy > 0; i ++){

        if(i < a.ln) cy += a.v[i];

        if(i < b.ln) cy += b.v[i];

        res.v[i] = cy % base;

        cy /= base;

    }

    res.ln = i;

    return res;

}



bint operator - (const bint& a, const bint& b){

    bint res;

    int i, cy = 0;

    for(res.ln = a.ln, i = 0; i < res.ln; i ++){

        res.v[i] = a.v[i] - cy;

        if(i < b.ln) res.v[i] -= b.v[i];

        if(res.v[i] < 0) cy = 1, res.v[i] += base;

        else cy = 0;

    }

    while(res.ln > 0 && res.v[res.ln - 1] == 0) res.ln --;

    return res;

}



bint operator * (const bint& a, const bint& b)

{

    bint res;res.ln = 0;

    if(0 == b.ln) {res.v[0] = 0; return res;}

    int i, j, cy;

    for(i = 0; i < a.ln; i ++){

        for(j = cy = 0; j < b.ln || cy > 0; j ++, cy /= base)

        {

            if(j < b.ln) cy += a.v[i] * b.v[j];

            if(i + j < res.ln) cy += res.v[i + j];

            if(i + j >= res.ln) res.v[res.ln ++] = cy % base;

            else res.v[i + j] = cy % base;

        }

    }

    return res;

}



int digits(bint& a)

{

    if(a.ln == 0) return 0;

    int l = (a.ln - 1) * 4;

    for(int t = a.v[a.ln - 1]; t; ++ l, t /= 10);

    return l;

}



bool read(bint& b, char buf[])

{

    if(1 != scanf("%s", buf)) return 0;

    int w, u, ln = strlen(buf);

    memset(&b, 0, sizeof(bint));

    if('0' == buf[0] && 0 == buf[1]) return 1;

    for(w = 1, u = 0; ln; ) {

        u += (buf[-- ln] - '0') * w;

        if(w * 10 == base){

            b.v[b.ln ++] = u; u = 0; w = 1;

        }

        else w *= 10;

    }

    if(w != 1) b.v[b.ln ++] = u;

    return 1;

}



void write(const bint & b){

    int i;

    printf("%d", b.ln == 0 ? 0 : b.v[b.ln - 1]);

    for(i = b.ln - 2; i >= 0; i --)

        printf("%04d", b.v[i]);

    printf("\n");

}



int main()

{

    char op[5];

    while(true)

    {

        if(!read(a, buf)) break;

        scanf("%s", op);

        memset(buf, 0, sizeof buf);

        read(b, buf);

        memset(&c, 0, sizeof(bint));

        if('+' == op[0])

        {

            c = a + b;

        }

        else if('-' == op[0])

        {

            c = a - b;

        }

        else if('*' == op[0])

        {

            c = a * b;

        }

        write(c);

    }

    return 0;

}

G：

// n * x - m * y = 1; d = gcd(n, m);



#include<stdio.h>

#include<string.h>

#include<stdlib.h>



typedef long long LL;

LL n, m, x, y;



LL extgcd( LL a, LL b, LL &x, LL &y)

{

    if( b == 0) { x = 1; y = 0; return a;}

    LL d = extgcd( b, a % b, x, y);

    LL t = x;

    x = y;

    y = t - a / b * y;

    return d;

}



int main()

{

    while(scanf("%lld%lld", &n, &m) == 2)

    {

        LL d = extgcd(n, m, x, y);

        LL t = 1;

        if( t % d != 0)

        {

            printf( "No Solution.\n");

        }

        else{

            x = t / d * x;

            x = ( x % (m / d) + (m / d)) % (m / d);

            printf( "%lld\n", x) ;

        }



    }

    return 0;

}

I：

#include<stdio.h>

#include<math.h>



const double eps = 1e-8;



double x1, yy1, x2, y2, x3, y3, x4, y4;

double k1, k2, b1, b2;

double x, y;



int main()

{

    while(scanf("%lf%lf%lf%lf", &x1, &yy1, &x2, &y2) != EOF)

    {

        scanf("%lf%lf%lf%lf", &x3, &y3, &x4, &y4);

        if(fabs(x1 - x2) < eps || fabs(x3 - x4) < eps)

        {

            if(fabs(x1 - x2) < eps && fabs(x3 - x4) < eps)

            {

                printf("No Solution!\n");

                continue;

            }

            if(fabs(x1 - x2) < eps)

            {

                k2 = (y3 - y4) / (x3 - x4);

                b2 = y3 - k2 * x3;

                y = x1 * k2 + b2;

                x = x1;

            }



            if(fabs(x3 - x4) < eps)

            {

                k1 = (yy1 - y2) / (x1 - x2);

                b1 = yy1 - k1 * x1;

                y = x3 * k1 + b1;

                x = x3;

            }

            printf("%.6f %.6f\n", x, y);

            continue;

        }

        else

        {

            k1 = (yy1 - y2) / (x1 - x2);

            b1 = yy1 - k1 * x1;

            k2 = (y3 - y4) / (x3 - x4);

            b2 = y3 - k2 * x3;

            if(fabs(k1 - k2) < eps)

            {

                printf("No Solution!\n");

                continue;

            }

            x = (b2 - b1) / (k1 - k2);

            y = k1 * x + b1;

            printf("%.6f %.6f\n", x, y);

        }

    }

    return 0;

}