20130906组队赛HDU4236-HDU4237-HDU4239-HDU4242-HDU4243

A -Repeating Characters

 

A题看样例就可以写了，直接重复就可以了：

 

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int QuickMod(int a, int b, int n);char num[50];int main(){    int t, n, ca;    scanf("%d", &t);    while(t--)    {        mem(num, 0);        scanf("%d%d%s", &ca, &n, num);        int len = strlen(num);        printf("%d ", ca);        for(int i = 0; i < len; ++i)        {            for(int j = 0; j < n; ++j)            printf("%c", num[i]);        }        printf("\n");    }    return 0;}int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}

B -The Rascal Triangle

 

B题有好几种推法的，我是直接从每一行的第一个开始推，对于第i行，它的第二个数字==1+（n-1）。第二个数==1+（n-1)+（n-3），就这样一直推就可以了，我写的是分成前后两部分算，后来发现直接不管前半部分还是后半部分都适用这个公式，一路算过去就可以了，一个for：

 

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define maxn 50001int QuickMod(int a, int b, int n);int main(){    int t, ca, n, m;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &ca, &n, &m);        printf("%d ", ca);        long long ans = 1;        if(m <= n/2)        {            int base = 1;            for(int i = 1; i <= m; ++i)            {                ans += (n-base);                base += 2;            }            printf("%lld\n", ans);        }        else        {            int cha = n - m - 1;            int base = 1;            for(int i = 0; i <= cha; ++i)            {                ans += (n-base);                base += 2;            }            printf("%lld\n", ans);        }    }    return 0;}int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}

D -Decoding EDSAC Data

 

D题麻烦的题目，我们一直在搞C题，C题虽然说是和D题是相反的，但是我们的方法在精度这块儿卡死了。好多D都

过了，就先去搞D。D题我们是直接一对一的把这些符号转换成二进制，再把二进制转换成小数（-1.0特判）。一直

没有对负数的二进制求补，所以不对，后来对负数的二进制求了补就出答案了。可是写的太少，在怎么输出这里卡

了好久，最后还是先确定了这个数字应该输出几位，用C++的输出搞定了：

 

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define ll long long#define eps 1e-20map <char , string> mm;void init(){    mm['P'] = "00000";    mm['Q'] = "00001";    mm['W'] = "00010";    mm['E'] = "00011";    mm['R'] = "00100";    mm['T'] = "00101";    mm['Y'] = "00110";    mm['U'] = "00111";    mm['I'] = "01000";    mm['O'] = "01001";    mm['J'] = "01010";    mm['#'] = "01011";    mm['S'] = "01100";    mm['Z'] = "01101";    mm['K'] = "01110";    mm['*'] = "01111";    mm['?'] = "10000";    mm['F'] = "10001";    mm['@'] = "10010";    mm['D'] = "10011";    mm['!'] = "10100";    mm['H'] = "10101";    mm['N'] = "10110";    mm['M'] = "10111";    mm['&'] = "11000";    mm['L'] = "11001";    mm['X'] = "11010";    mm['G'] = "11011";    mm['A'] = "11100";    mm['B'] = "11101";    mm['C'] = "11110";    mm['V'] = "11111";}int main(){    init();    int T, ca, x;    char op1, op2;    cin >> T;    while(T--)    {        cin >> ca >> op1 >> x >> op2;        printf("%d ",ca);        if(op1 == '?' && x == 0 && op2 == 'F')        {            printf("-1.0\n");            continue;        }        string tmp = "";        tmp = mm[op1];        string s;        for(int i = 15; i >= 5; i--)        {            if(x%2 == 0)                s += '0';            else                s += '1';            x /= 2;        }        reverse(s.begin(), s.end());        tmp += s;        if(op2 == 'F')            tmp += '0';        else            tmp += '1';        int flag=0;        if(tmp[0] == '1')        {            flag = 1;            int k;            for(int i = 16; i >= 1; i--)            {                if(tmp[i]=='1')                {                    k = i;                    break;                }            }            k--;            for(int i = 1; i <= k; i++)            {                if(tmp[i]=='1')                    tmp[i] = '0';                else tmp[i] = '1';            }        }        int pos = 1;        for(int i=16; i>=1; i--)        {            if(tmp[i] != '0')            {                pos = i;                break;            }        }        double ans=0,k=0.5;        for(int i = 1; i <= pos; i++)        {            if(tmp[i]=='1')                ans+=k;            k /= 2;        }        if(flag == 1)            putchar('-');        cout << setiosflags(ios::fixed) << setprecision(pos) << ans << endl;    }    return 0;}

 

 

G -Rancher's Gift

 

这道题目是一道几何题目，比赛的时候没看懂题目，纠结了就没有做，后来看了，额，就是赤裸裸的计算，给出

A，B，C, D的坐标求出这五个多变形的面积和中间那个四边形的周长。刚刚A掉了，我的方法比较烦，求出了四个

交点的坐标，推导的时候比较繁一点，推出来就是套公式了。剩下的就是算面积而已。分成三角形来算。水了

 

代码：

 

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define ll long long#define eps 1e-20double findx(double ax, double ay, double bx, double by, double cx, double cy, double dx, double dy){    double fz = (bx - ax)*(cx*dy - cy*dx) - (dx - cx)*(ax*by - ay*bx);    double fm = dy*(bx - ax) - cy*(bx - ax) - by*(dx - cx) - ay*(cx - dx);    return fz/fm;}double findy(double x, double ax, double ay, double bx, double by){    double fz = (x - ax)*(by - ay);    double fm = (bx - ax);    return (fz/fm) + ay;}double dis(double x1, double y1, double x2, double y2){    return sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));}double sum(double x1, double y1, double x2, double y2, double x3, double y3){    double len1 = dis(x1, y1, x2, y2);    double len2 = dis(x1, y1, x3, y3);    double len3 = dis(x2, y2, x3, y3);    double S = (len1+len2+len3)/2.0;    return sqrt(S * (S - len1) * (S - len2) * (S - len3));}int main(){    double ax, ay, bx, by, cx, cy, dx, dy;    ax = 0;    ay = 0;    by = 0;    double abx, aby, bcx, bcy, cdx, cdy, dax, day;    double x1, y1, x2, y2, x3, y3, x4, y4;    int t, num;    scanf("%d", &t);    while(t--)    {        scanf("%d%lf%lf%lf%lf%lf", &num, &bx, &cx, &cy, &dx, &dy);        abx = (ax + bx)/2.0;        aby = (ay + by)/2.0;        bcx = (bx + cx)/2.0;        bcy = (by + cy)/2.0;        cdx = (cx + dx)/2.0;        cdy = (cy + dy)/2.0;        dax = (dx + ax)/2.0;        day = (dy + ay)/2.0;        x1 = findx(ax, ay, bcx, bcy, dx, dy, abx, aby);        y1 = findy(x1, ax, ay, bcx, bcy);        x2 = findx(ax, ay, bcx, bcy, bx, by, cdx, cdy);        y2 = findy(x2, ax, ay, bcx, bcy);        x3 = findx(cx, cy, dax, day, bx, by, cdx, cdy);        y3 = findy(x3, cx, cy, dax, day);        x4 = findx(cx, cy, dax, day, dx, dy, abx, aby);        y4 = findy(x4, cx, cy, dax, day);//        cout << x1 <<' ' << y1 << endl;//        cout << x2 <<' ' << y2 << endl;//        cout << x3 <<' ' << y3 << endl;//        cout << x4 <<' ' << y4 << endl;        double s1, s2, s3, s4, ss;        s1 = sum(ax, ay, bx, by, x2, y2)/160;        s2 = sum(bx, by, cx, cy, x3, y3)/160;        s3 = sum(cx, cy, dx, dy, x4, y4)/160;        s4 = sum(ax, ay, dx, dy, x1, y1)/160;        ss = (sum(x1, y1, x2, y2, x3, y3) + sum(x1, y1, x3, y3, x4, y4))/160;        int lensum = (int)ceil((dis(x1, y1, x2, y2) + dis(x2, y2, x3, y3) + dis(x3, y3, x4, y4) + dis(x4, y4, x1, y1)) * 16.5);        printf("%d %.3lf %.3lf %.3lf %.3lf %.3lf %d\n", num, s1, s2, s3, s4, ss, lensum);    }}

 

H -Maximum in the Cycle of 1

 

H题就是一个排列组合的问题，对于n,对于找到1之前的最大的数字K，求这样的序列有多少个。我们分情况讨论如

果是1的话，那么就是（n-1）！。然后当k大于1的时候，我们就求和，1-k-1一步的情况，然后1-x-k-1, 1-k-x-1

两步的情况，1-x-y-k-1,1-x-k-y-1,1-y-x-k-1,1-y-k-x-1，三步的情况，就这样一直加下去就可以了，其余的数

字全排列就行，就是i步的情况A[i]*C[k-2][i-1]*A[n-i-1]；

 

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;#define ll long longll A[19];ll C[19][19];void init(){    A[0] = 1;    A[1] = 1;    for(int i = 2; i < 19; ++i)        A[i] = i*A[i-1];    C[0][0] = 1;    C[1][0] = 1;    C[1][1] = 1;    for(int i = 2; i < 19; ++i)        for(int j = 0; j <= i; ++j)            C[i][j] = (A[i]/((A[i-j]*A[j])));//    for(int i = 0; i < 19; ++i)//        cout << A[i] << endl;//    for(int i = 0; i < 19; ++i)//    {//        for(int j = 0; j <= i; ++j)//            cout << C[i][j] <<' ';//        cout << endl;//    }}int main(){    init();    int t, ca, n, k;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &ca, &n, &k);        ll ans = 0;        printf("%d ", ca);        if(k == 1)        {            printf("%lld\n", A[n-1]);            continue;        }        for(int i = 1; i <= k-1; ++i)            ans += (A[i]*C[k-2][i-1]*A[n-i-1]);        printf("%lld\n", ans);    }    return 0;}