2019.10.13考试解题报告

2019.10.13考试解题报告

总结

期望得分：(100 + 30 + 0)
实际得分：(100 + 50 + 0)

神奇的多得了(20)分？
花两个多小时去推(T1)的式子，结果是推出来了，慌慌张张差点翻车
去打(T2)的暴力，顺便打了几个(if)，然后就(30-->50)?心里有点小激动
(T3)没时间做了……我菜

---

思路

T1

一开始就想写正解，发现不会，于是打了个(40)分的暴力保底，对暴力的式子进行拆解和优化，得到了下面的过程

要求的是

[min sum_{j = 1}^{n} abs(a_i - a_j) * b_j ]

首先去掉绝对值号，发现我们能把这个式子分成两部分

[ ans +=left{ egin{aligned} sum_{j = 1}^{i}(a_i - a_j) * b_j (a_i > a_j) \ sum_{j = i}^{n}(a_j - a_i) * b_j (a_i < a_j) end{aligned} ight. ]

然后拆解发现，前一部分变成了

[a_ib_1 - a_1b_1 + a_ib_2 - a_2b_2 + ……a_ib_j-a_jb_j ]

后一部分变成了

[a_jb_j-a_ib_j+a_{j+1}b_{j+1} -a_ib_{j+1}+……a_nb_n-a_ib_n ]

然后在合并一下子（我懒得写了）

然后发现我们可以用前缀和！！！

然后就做出来了！！

T2

(30)分暴力很好打，暴搜就行了

(60)的话，分别枚举两个骰子点数，计算出来每种点数和的方案数（除以(6^n)得到概率）
如果你扔出了(i)点，那你胜利的方案是对手扔出([1,i))，从小到大枚举你的点数，同时
计算对手的前缀和

满分做法
计算每种点数的概率
计算(i)个骰子的概率时，考虑最后一个骰子的点数为(1-6)的每种情况，可以从(i-1)个骰
子概率转移过来。
记(f[i][j])表示(i)个骰子扔出(j)的概率
(f[i][j] = sum{f[i-1][j-k]/6}(1≤k≤6))
递推即可，复杂度(O(n²))

T3

神仙题神仙题神仙题

---

代码

T1

考场满分

//By:Loceaner
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 1e6 + 11;

struct node {
	int a, b, id, c;
} e[N];
 
bool cmp(node x, node y) {
	return x.a < y.a;
}

int n, sum2[N], sum3[N], ans = 0x3f3f3f3f3f3f3f3f;
bool flag = 1;

signed main() {
	n = read();
	for(int i = 1; i <= n; i++) {
		e[i].a = read();
		if(i != 1)
			if(e[i].a != e[i - 1].a) 
				flag = 0;
	}
	for(int i = 1; i <= n; i++) e[i].b = read(), e[i].c = e[i].a * e[i].b;
	if(flag) return cout << "0
", 0;
	stable_sort(e + 1, e + 1 + n, cmp);
	for(int i = 1; i <= n; i++) {
		sum2[i] = sum2[i - 1] + e[i].b;
		sum3[i] = sum3[i - 1] + e[i].c;
	}
//	for(int i = 1; i <= n; i++) {
//		int now = 0;
//		for(int j = 1; j <= n; j++) {
//			now += abs(e[i].a - e[j].a) * e[j].b;
//		}
//		ans = min(ans, now);
//	}
	int now = 0;
	for(int i = 1; i <= n; i++) {
		now = e[i].a * sum2[i - 1] - sum3[i - 1] + (sum3[n] - sum3[i]) - e[i].a * (sum2[n] - sum2[i]);
		ans = min(ans, now);
	}
	cout << ans << '
'; 
	return 0;
}
/*
6
2 1 3 3 2 4
5 1 1 2 2 2
10
1 2 2 5 6 6 8 10 10 10
9 55 41 2 4 6 3 2 4 10
*/

题解里的正解

#include<bits/stdc++.h>
#define lson (o<<1)
#define rson (o<<1|1)
#define fi first
#define sc second
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
using namespace std;
const double pi=acos(-1);
const double eps=1e-6;
inline int lowbit(int x){return x&(-x);}
inline int read(){
    int f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
template<typename T> inline T max(T x,T y,T z){return max(max(x,y),z);}
template<typename T> inline T min(T x,T y,T z){return min(min(x,y),z);}
template<typename T> inline T sqr(T x){return x*x;}
template<typename T> inline void checkmax(T &x,T y){x=max(x,y);}
template<typename T> inline void checkmin(T &x,T y){x=min(x,y);}
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B,typename C> inline A fpow(A x,B p,C yql){
    A ans=1;
    for(;p;p>>=1,x=1LL*x*x%yql)if(p&1)ans=1LL*x*ans%yql;
    return ans; 
}
struct FastIO{
    static const int S=1310720;
    int wpos;char wbuf[S];
    FastIO():wpos(0) {}
    inline int xchar(){
        static char buf[S];
        static int len=0,pos=0;
        if(pos==len)pos=0,len=fread(buf,1,S,stdin);
        if(pos==len)return -1;
        return buf[pos++];
    }
    inline int read(){
        int c=xchar(),x=0;
        while(c<=32&&~c)c=xchar();
        if(c==-1)return -1;
        for(;'0'<=c&&c<='9';c=xchar())x=x*10+c-'0';
        return x;
    }
}io;
//#define read io.read
const int N=1e6+10;
const int yql=1e9+7;
struct Node{
    int x,v;
}a[N];
inline bool cmp(Node a,Node b){return a.x<b.x;}
int n,m;
ll cur=0,ans=0,k1,k2; 
int main(){
    freopen("shiroha1.in","r",stdin);
    freopen("shiroha1.out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++)a[i].x=read();
    for(int i=1;i<=n;i++)a[i].v=read();
    sort(a+1,a+n+1,cmp);
    for(int i=1;i<=n;i++)cur=cur+1LL*(a[i].x-a[1].x)*a[i].v;
    ans=cur;
    for(int i=2;i<=n;i++)k1+=a[i].v;k2=a[1].v;
    for(int i=2;i<=n;i++){
        //dbg(cur);
        //cur=(cur-1LL*k1*(a[i].x-a[i-1].x)%yql+yql)%yql;
        //cur=(cur+1LL*k2*(a[i].x-a[i-1].x)%yql)%yql;
        cur=cur-1LL*k1*(a[i].x-a[i-1].x);
        cur=cur+1LL*k2*(a[i].x-a[i-1].x);
        k1-=a[i].v;k2+=a[i].v;
        checkmin(ans,cur);
    }
    printf("%lld
",ans);
}

T2

暴力

//By:Loceaner
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

int x, y, nowx, nowy, cnta, cntb, cntp;

void dfs2(int tot, int ans) {
	if(tot == y) {
		nowy = ans;
		if(nowx == nowy) cntp++;
		else if(nowx > nowy) cnta++;
		else cntb++;
		return;
	}
	for(int i = 1; i <= 6; i++) {
		dfs2(tot + 1, ans + i);
	}
}

void dfs1(int tot, int ans) {
	if(tot == x) {
		nowx = ans;
		dfs2(0, 0);
		return;
	}
	for(int i = 1; i <= 6; i++) {
		dfs1(tot + 1, ans + i);
	}
}

int main() {
	x = read(), y = read();
	if(x + y <= 10) {
		dfs1(0, 0);
		printf("%.2lf%%", ((1.0 * cnta) / (cnta + cntb + cntp)) * 100);
	} 
	else if(x - 7 >= y) cout << "100.00%";
	else if(y - 7 >= x) cout << "0.00%";
	else if(x > y * 6) cout << "100.00%";
	else if(x == y) cout << "50%";
	else cout << "0.00%";
	return 0;
}

正解

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

double a[1001][7001], sx, sy, ans;
int x, y;

int main() {
	scanf("%d%d", &x, &y);
	int xy = x > y ? x : y;
	for(int i = 1; i <= 6; i++) a[1][i] = 1;
	for(int i = 2; i <= xy; i++)
		for(int j = i; j <= 6 * i; j++)
			for(int k = 1; k <= 6; k++)
				if(j > k) a[i][j] += a[i - 1][j - k] / 6.0;
	for(int i = x; i <= 6 * x; i++) {
		for(int j = y; j <= 6 * y; j++) 
			sy += a[x][i] * a[y][j];
		if(i <= y) continue;
		for(int j = y; j <= i - 1; j++) 
			sx += a[y][j] * a[x][i];
	}
	ans = sx / sy * 100;
	printf("%.2lf%%", ans);
	return 0;
}

T3

正解

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN = 1000010;
const int MAXL = 22;

int rmin[MAXL][MAXN], rmax[MAXL][MAXN];

int n, q;
int a[MAXN];

inline void init_rmq() {
    for (int i = 1; i < MAXL; i++)  {
        for (int j = 0; j + (1 << i) <= n; j++) {
            rmin[i][j] = min(rmin[i-1][j], rmin[i-1][j+(1<<(i-1))]);
            rmax[i][j] = max(rmax[i-1][j], rmax[i-1][j+(1<<(i-1))]);
        }
    }
}

inline bool test(int d, int m) {
    int l = 0, i;
    for (i = 0; i < m && l < n; i++) {
        int r = l, smin = 1000000001, smax=0;
        for (int j = MAXL - 1; j >= 0; j--) {
            if (r + (1 << j) <= n) {
                int tmin = rmin[j][r], tmax = rmax[j][r];
                if (max(smax, tmax) - min(smin, tmin) <= d) {
                    smax = max(smax, tmax);
                    smin = min(smin, tmin);
                    r += (1 << j);
                }
            }
        }
        l = r;
    }
    return l == n;
}

inline int work(int m) {
    int l = 0, r = 1000000001;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (test(mid, m)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%*d%d", &rmin[0][i]);
        rmax[0][i] = rmin[0][i];
    }
    init_rmq();
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
        int m;
        scanf("%d", &m);
        int ans = work(m);
        if (ans % 2) printf("%d.5
", ans/2);
        else printf("%d
", ans/2);
    }
}