2018 ACM 国际大学生程序设计竞赛上海大都会部分题解

题目链接

2018 ACM 国际大学生程序设计竞赛上海大都会

下午午休起床被同学叫去打比赛233

然后已经过了2.5h了

先挑过得多的做了

....

A题

rand x*n 次点，每次judge一个点位端点的共线向量数判断是否大于给定x

强行rand 500次

代码

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x = 0,f = 1;
    char c = getchar();
    while(c < '0' || c > '9') {if(c == '-')f =- 1;  c = getchar(); }
    while(c <= '9' &&c >= '0') x = x * 10 + c - '0',c = getchar();
    return x * f;
}
const int maxn = 20007;
#define eps (1e-13)
double k;
int n;
struct points {
    int x,y;
} p[maxn];
bool solve(int p1,int p2) {
    int tx = p[p1].x - p[p2].x,ty = p[p1].y - p[p2].y;
    int cnt = 2;
    for(int i = 1;i <= n;++ i) {
        if(i == p1 || i == p2) continue;
        int px = p[p1].x - p[i].x,py = p[p1].y - p[i].y;
        if((long long)tx * py - (long long)ty * px == 0) cnt ++;
    }
    double K = (double) cnt /  (double) n;
    return (K >= k);
}
int main() {
    srand(time(0));
    srand(rand());
    int t = read();
    while(t -- ) {
        n = read(); scanf("%lf",&k);
        for(int i = 1;i <= n;++ i) p[i].x = read(),p[i].y = read();
        int p1,p2;
        bool flag = false;
        for(int i = 500;i --;) {
            p1 = rand() % n + 1;
            p2 = p1;
            while(p2 == p1) p2 = rand() % n + 1;
            if(solve(p1,p2)) {
                flag = true; break;
            }
            if(flag) break;
        }
        if(t != 0) {
            if(flag) puts("Yes");
            else puts("No");
        }
        if(t == 0) {
            if(flag)printf("Yes");
            else printf("No");
        }
    }
    return 0;
}

B题

很sb的打了nlogn筛

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 200000;
vector<int>vec[maxn + 10];
int num[maxn  + 10]; 

int main() {
    for(int i = 1;i <= maxn;++ i) {
        for(int j = i;j <= maxn;j += i) {
            num[j] += i;vec[j].push_back(i);
        }
    }
    int t;
    scanf("%d",&t);
    int cas = 0;
    while(t --) {
        cas++;
        int k ;
        scanf("%d",&k); printf("Case %d: ",cas);
        if(num[k] - k != k) {
            if(k != 0)puts("Not perfect.");
            else printf("Not perfect.");
        } else {
            int p = vec[k].size() - 1;
            printf("%d = ",k);
            for(int i = 0;i < p;++ i) {
                if(i != p - 1)printf("%d + ",vec[k][i]);
                else printf("%d",vec[k][i]);
            }
            if(t != 0)puts("");
        }
    }
    return 0;
}

D题Thinking-Bear magic

发现每次面积变为原来的$\frac{1}{\cos(\frac{\pi}{n})^2} $

代码

#include<bits/stdc++.h>
#define Pi 3.14159265359 

double Sin(int n){
    return sin((2.0*Pi)/n);
}
double Cos(int n){
    return cos((2.0*Pi)/n);
}
double Tan(int n){
    return tan((2.0*Pi)/n);
}
double S(int n, double R) {
    return ((0.25*n*R*R)/(double)tan(Pi/n));
}
double calc(int n){
    return cos(Pi/(1.0*n))*cos(Pi/(1.0*n));
}
int main () {
    int T,n,a,l;
    scanf("%d",&T);
    double s;
    int t;
    while(T --) {
        scanf("%d%d%d",&n,&a,&l);
        double gg = calc(n);
        t = 0; s = S(n,(double)a);
        while(s > l) s = gg * s, ++ t;
        printf("%d\n",t);
    }
    return 0;
}

J题目 Beautiful Numbers

数位dp

代码

#include<bits/stdc++.h>
using namespace std;
inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
const int maxn=3e5+5;
const int INF =0x3f3f3f3f;
#define LL long long
int a[20];
LL dp[13][105][150];
int mod;
LL dfs(int pos,int sum,int res,bool limit) {
    if(pos == -1) return (sum == mod && !res);
    if(dp[pos][sum][res] != -1 && !limit) return dp[pos][sum][res];
    int up = limit ? a[pos] : 9;
    LL res = 0;
    for(int i = 0;i <= up;++ i) {
        if(i + sum > mod) break;
        res += dfs(pos - 1,sum + i,(res * 10 + i) % mod,limit && i == a[pos]);
    }
    if(!limit) dp[pos][sum][res] = res;
    return res;
}
LL solve(LL n) {
    int pos=0;
    LL x = n;
    while(x) a[pos ++] = x % 10, x /= 10;
    LL res = 0;
    for(int i = 1;i <= 9 * pos;++ i) {
        mod = i;
        memset(dp,-1,sizeof(dp));
        res += dfs(pos-1,0,0,true);
    }
    return res;
}
int main() {
    int T = read();
    int cas = 0;
    while(T --) {
        cas ++;
        LL n = read();
        printf("Case %d: %lld\n",cas,solve(n));
    }
    return 0;
}

K题 Matrix Multiplication

矩乘裸题

代码

#include<bits/stdc++.h>
using namespace std;
struct MMMM {
    int m[21][21];
    int r,c;
    MMMM() {}
    MMMM(int n) {memset(m, false, sizeof m);r = c = n; for (int i = 0; i < r; i +=1) m[i][i] = 1;}
    MMMM(int R,int C) {memset(m, false, sizeof m);r = R,c = C;}
    MMMM operator * (const MMMM & o)const {
        MMMM res = MMMM(r, o.c);
        for (int i = 0; i< r;i += 1)
            for (int  j = 0 ;j < o.c; j += 1)
                for (int k = 0; k < c; k += 1 )
                    res.m[i][j] += 1ll * m[i][k] * o.m[k][j];
        return res;
    }
    void print() {
        for (int i =0 ;i < r;i += 1) {
            for (int j = 0; j < c; j += 1) {
                if(j!=c-1)printf("%d ",m[i][j]);
                else printf("%d",m[i][j]);
            }
            puts("");
        }
    }
    void input() {
        for (int i = 0; i< r; i += 1)
            for (int j = 0; j < c; j += 1)
                scanf("%d", &m[i][j]);
    }
};
int main () {
    int T;
    scanf("%d",&T);
    for (int i =1; i <= T; i += 1) {
        int r1,c1,r2,c2;
        scanf("%d%d%d%d", &r1,&c1,&r2,&c2);
        MMMM a,b;
        a=MMMM(r1,c1), b= MMMM(r2,c2);
        a.input(); b.input();
        printf("Case %d:\n",i);
        if (c1 != r2) {
            printf("ERROR\n");
            continue;
        }
        MMMM c = a * b;
        c.print();
    }
    return 0;
}