2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

02Train Seats Reservation

33.87%
1000ms
131072K

You are given a list of train stations, say from the station

The passengers can order several tickets from one station to another before the train leaves the station one. We will issue one train from the station

Note that one single seat can be used by several passengers as long as there are no conflicts between them. For example, a passenger from station

Input Format

Several sets of ticket reservations. The inputs are a list of integers. Within each set, the first integer (in a single line) represents the number of orders,

Output Format

For each set of ticket reservations appeared in the input, calculate the minimum number of seats required so that all reservations are satisfied without conflicts. Output a single star '*' to signify the end of outputs.

样例输入

样例输出

20
60
*

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个B是我的暴力扫描线做得不太好，我承认，我是个sb，本来卡long long，后来卡扫描线，可以先下后上嘛，所以是左闭右开区间

#include <bits/stdc++.h>
using namespace std;
const int N=1005;
int n,a[N],b[N];
int main()
{
    while(~scanf("%d",&n))
    {
        if(!n)
        {
            printf("*
");
            break;
        }
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0; i<n; i++)
        {
            int l,r,x;
            scanf("%d%d%d",&l,&r,&x);
            a[l]+=x;
            b[r]+=x;
        }
        long long f=0,ma=0;
        for(int i=1; i<=100; i++)
        {
            f-=b[i];
            f+=a[i];
            ma=max(f,ma);
        }
        printf("%lld
",ma);
    }
    return 0;
}

问答

37.67%
1000ms
131072K

There are

Although the problem looks simple at the first glance, it might take a while to figure out how to do it correctly. Note that the shape of the union can be very complicated, and the intersected areas can be overlapped by more than two rectangles.

Note:

(1) The coordinates of these rectangles are given in integers. So you do not have to worry about the floating point round-off errors. However, these integers can be as large as

(2) To make the problem easier, you do not have to worry about the sum of the areas exceeding the long integer precision. That is, you can assume that the total area does not result in integer overflow.

Input Format

Several sets of rectangles configurations. The inputs are a list of integers. Within each set, the first integer (in a single line) represents the number of rectangles, n, which can be as large as

These configurations of rectangles occur repetitively in the input as the pattern described above. An integer

Output Format

For each set of the rectangles configurations appeared in the input, calculate the total area of the union of the rectangles. Again, these rectangles might overlap each other, and the intersecting areas of these rectangles can only be counted once. Output a single star '*' to signify the end of outputs.

样例输入

样例输出

7
3
*

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个F网上直接有代码啊，直接拉过看来就行的，暴力扫描线肯定不行。线段树搞一下

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ls i<<1
#define rs i<<1|1
#define m(i) ((q[i].l + q[i].r)>>1)
using namespace std;
typedef long long ll;
const int N = 1001;
struct Edge
{
    double l,r;
    double h;
    int f;
}e[N<<1];
bool cmp(Edge a,Edge b)
{
    return a.h < b.h;
}
struct Node
{
    int l,r;
    int s;
    double len;
}q[N*8];
double x[2*N];
void build(int i,int l,int r)
{
    q[i].l = l,q[i].r = r;
    q[i].s = 0;q[i].len = 0;
    if (l == r) return;
    int mid = m(i);
    build(ls,l,mid);
    build(rs,mid+1,r);
}
void pushup(int i)
{
    if (q[i].s)
    {
        q[i].len = x[q[i].r+1] - x[q[i].l];
    }
    else if (q[i].l == q[i].r)
    {
        q[i].len = 0;
    }
    else
    {
        q[i].len = q[ls].len + q[rs].len;
    }
}
void update(int i,int l,int r,int xx)
{ 
    if (q[i].l == l&&q[i].r == r)
    {
        q[i].s += xx;
        pushup(i);
        return;
    }
    int mid = m(i);
    if (r <= mid) update(ls,l,r,xx);
    else if (l > mid) update(rs,l,r,xx);
    else
    {
        update(ls,l,mid,xx);
        update(rs,mid+1,r,xx);
    }
    pushup(i);
}
int main()
{
    int n;int kas = 0;
    while (scanf("%d",&n),n)
    {
        int tot = 0;
        for (int i = 0;i < n;++i)
        {
            double x1,x2,y1,y2;
            scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
            Edge &t1 = e[tot];Edge &t2 = e[1+tot];
            t1.l = t2.l = x1,t1.r = t2.r = x2;
            t1.h = y1;t1.f = 1;
            t2.h = y2;t2.f = -1;
            x[tot] = x1;x[tot+1] = x2;
            tot += 2;
        }
        sort(e,e+tot,cmp);
        sort(x,x+tot);
        int k = 1;
        for (int i = 1;i < tot;++i)
        {
            if (x[i] != x[i-1])
            {
                x[k++] = x[i];
            }
        }
        build(1,0,k-1);
        double ans = 0.0;
        for (int i = 0;i < tot;++i)
        {
            int l = lower_bound(x,x+k,e[i].l) - x;
            int r = lower_bound(x,x+k,e[i].r) - x - 1;
            update(1,l,r,e[i].f);
            ans += (e[i+1].h - e[i].h)*q[1].len;
        }
        printf("%.0f
",ans);
    }
    puts("*");
}

问答

41.93%
1000ms
262144K

In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph.

Given an integer $S_{n}Sn, is an undirected graph consisting of n!n! nodes (or vertices) and ((n-1) * n!)/2((n−1) ∗ n!)/2 edges. Each node is uniquely assigned a label x_{1} x_{2} ... x_{n}x1 x2 ... xnwhich is any permutation of the n digits {1, 2, 3, ..., n}1,2,3,...,n. For instance, an S_{4}S4 has the following 24 nodes {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321. For each node with label x_{1} x_{2} x_{3} x_{4} ... x_{n}x1 x2x3 x4 ... xn, it has n-1n−1 edges connecting to nodes x_{2} x_{1} x_{3} x_{4} ... x_{n}x2 x1 x3 x4 ... xn, x_{3} x_{2} x_{1} x_{4} ... x_{n}x3 x2 x1 x4 ... xn, x_{4} x_{2} x_{3} x_{1} ... x_{n}x4 x2x_{n} x_{2} x_{3} x_{4} ... x_{1}n-1d-thx_{1} x_{2} x_{3} x_{4} ... x_{n}d = 2, ..., nS_{4}12343213432144231S_{4}abcd x3 x1 ... xn, ..., and xn x2 x3 x4 ... x1. That is, the n−1 adjacent nodes are obtained by swapping the first symbol and the d−th symbol of x1 x2 x3 x4 ... xn, for d=2,...,n. For instance, in S4, node 1234has 3 edges connecting to nodes 2134, 3214, and 4231. The following figure shows how S4looks (note that the symbols a, b, c, and d are not nodes; we only use them to show the connectivity between nodes; this is for the clarity of the figure).$

In this problem, you are given the following inputs:

Two nodes $x_{1}x1 x_{2}x2 x_{3}x3 ... x_{n}xn and y_{1}y1 y_{2}y2 y_{3} ... y_{n}y3 ... yn in S_{n}Sn.$

You have to calculate the distance between these two nodes (which is an integer).

Input Format

A list of

Output Format

A list of

样例输入

样例输出

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个题也很水，bfs一下就好，因为9！这个数并不大

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<queue>
#define LL long long
using namespace std;

struct mm
{
    int v,bu;
};
void bfs(int x,int y,int n)
{
    int b[10],k=0,i;    
    map<int ,int>a;
    a[x]=1;
    queue<mm>s;
    mm p;
    p.v=x;
    p.bu=0;
    s.push(p);
    while(!s.empty()){
        mm q=s.front();s.pop();
        int now=q.v,bushu=q.bu;
        if(q.v==y){
            printf("%d
",bushu);
            break;
        }
        k=n-1;
        while(now){
            b[k--]=now%10;
            now/=10;
        }
        for(i=1;i<n;i++){
            int sum=0,j,t,tt;
            t=b[i];
            tt=b[0];
            b[0]=t;
            b[i]=tt;
            for(j=0;j<n;j++){
                sum=sum*10+b[j];
            }
            if(!a[sum]){
                a[sum]=1;
                q.v=sum;
                q.bu=bushu+1;
                s.push(q);
            }
            b[0]=tt;
            b[i]=t;    
        }
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    int x,y;
    while(~scanf("%d%d",&x,&y)){
        bfs(x,y,n);

问答

34.36%
1000ms
131072K

Let $s_{1}s1, s_{2}s2, ......, s_{n}snEach integer is is associated with a weight by the following rules:$

(1) If is is negative, then its weight is

(2) If is is greater than or equal to $s_{i}si is s_{i}-10000si−10000 . For example, if s_{i}siis 1010110101, then is is reset to 101101 and its weight is 55.$

(3) Otherwise, its weight is

A non-decreasing subsequence of $s_{i1}si1, s_{i2}si2, ......, s_{ik}sik, with i_{1}<i_{2} ... <i_{k}i1<i2 ... <ik, such that, for all 1 leq j<k1≤j<k, we have s_{ij}<s_{ij+1}sij<sij+1.$

A heaviest non-decreasing subsequence of

Write a program that reads a sequence of integers, and outputs the weight of its

heaviest non-decreasing subsequence. For example, given the following sequence:

The heaviest non-decreasing subsequence of the sequence is

We guarantee that the length of the sequence does not exceed $2*10^{5}2∗105$

Input Format

A list of integers separated by blanks: $s_{1}s1, s_{2}s2,......,s_{n}sn$

Output Format

A positive integer that is the weight of the heaviest non-decreasing subsequence.

样例输入

80 75 73 93 73 73 10101 97 -1 -1 114 -1 10113 118

样例输出

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

转换一下，求最长不降子序列就可以的

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int N=1e6+10;
const int INF=0x3f3f3f3f;
int a[N],g[N],f[N],n;
int main() {
    int i=0,j,k;
    while(scanf("%d",&a[i])!=EOF)
    {
        if(a[i]>=10000)
        {
            j=0;
            k=a[i]-10000;
            while(j<5)
            {
                a[i]=k;
                j++;
                i++;
            }
        }
        else if(a[i]>=0)
        i++;
    }
        n=i;/*

    for(i=0;i<n;i++)printf("%d ",a[i]);*/
        fill(g,g+n,INF);
        for(int i=0; i<n; i++) {
            int j=lower_bound(g, g+n,a[i])-g;
            while(g[j]==a[i])
            j++;
            g[j]=a[i];
        }
        printf("%d
",lower_bound(g, g+n,INF)-g);
}

问答

46.98%
1000ms
131072K

The frequent subset problem is defined as follows. Suppose $S_{1}S1, S_{2}S2,ldots…,S_{M}SM are MM sets over UU. Given a positive constant alphaα, 0<alpha leq 10<α≤1, a subset BB (B eq 0B≠0) is α-frequent if it is contained in at least alpha MαM sets of S_{1}S1, S_{2}S2,ldots…,S_{M}SM, i.e. left | left { i:Bsubseteq S_{i} ight } ight | geq alpha M∣{i:B⊆Si}∣≥αM. The frequent subset problem is to find all the subsets that are α-frequent. For example, let U={1, 2,3,4,5}U={1,2,3,4,5}, M=3M=3, alpha =0.5α=0.5, and S_{1}={1, 5}S1={1,5}, S_{2}={1,2,5}S2={1,2,5}, S_{3}={1,3,4}S3={1,3,4}. Then there are 33 α-frequent subsets of UU, which are {1}{1},{5}{5} and {1,5}{1,5}.$

Input Format

The first line contains two numbers $S_{i}Si, 1 le i le M1≤i≤M . Your program should be able to handle NN up to 2020 and MM up to 5050.$

Output Format

The number of

样例输入

15 0.4
1 8 14 4 13 2
3 7 11 6
10 8 4 2
9 3 12 7 15 2
8 3 2 4 5

样例输出

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    long long n,i,j,b[51],l,x,m,y,s,h;
    double a,o;
    char c[10000];
    scanf("%lld%lf",&n,&a);
    j=0;
    memset(b,0,sizeof(b));
    getchar();
    while(gets(c)!=NULL)
    {
        l=strlen(c);
        i=0;
        while(i<l)
        {
            if(c[i]>='0'&&c[i]<='9')
            {
                x=0;
                while(i<l&&c[i]>='0'&&c[i]<='9')
                {
                    x=x*10+c[i]-'0';
                    i++;
                }
                b[j]|=(1<<x);
            }
            i++;
        }
        j++;
    }
    m=j;s=0;o=m*a;
    /*
for(i=0;i<m;i++)
    printf("%lld
",b[i]);*/
    for(i=1;i<(1<<(n+1));i++)
    {
        h=0;
        x=i;
        for(j=0;j<m;j++)
        {
            y=x&b[j];
            if(x==y)
            h++;
        }
        if(h>=o)
        s++;
    }
    printf("%lld
",s);
}

问答

23.24%
1000ms
131072K

Cache memories have been used widely in current microprocessor systems. In this problem, you are asked to write a program for a cache simulator. The cache has the following metrics:

1. The cache size is 1 KB (K-byte).

2. The cache uses the direct mapped approach.

3. The cache line size is 16 bytes.

4. The cacheable memory size is 256MB.

Your program will report a hit or miss when an address is given to the cache simulator. This is often called trace simulation. Initially, all of the cache lines are in the invalid state. When a memory line is first brought into the cache, the allocated cache entry transits into the valid state. Assume that a miss causes the filling of a whole cache line.

Input Format

Up to

Output Format

Report either

样例输入

AAAA000
00010B2
00010BA
END

样例输出

Miss
Miss
Hit
Hit ratio = 33.33%

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个H，毒瘤。。。cache

其实根据题目条件找到映射就可以了

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int a[65];
int main()
{
    char s[15];
    ios::sync_with_stdio(false);
    int m=0,m0=0;
    memset(a,-1,sizeof(a));
    while(cin>>s)
    {
        if(strcmp(s,"END")==0)break;
        m++;
        int n;
        sscanf(s,"%X",&n);
        int id=(n/16)%64,num=(n/16)/64;
        if(a[id]!=num)
        {
            a[id]=num;
            printf("Miss
");
        }
        else
        {
            printf("Hit
");
            m0++;
        }
    }
    printf("Hit ratio = %.2f%%
",m0*100.0/m);
    return 0;
}

问答

76.25%
1000ms
131072K

Consider a system which is described at any time as being in one of a set of $a_{ij}=p=[s_{i}=j | s_{i-1}=i], 1le i,j le Naij=p=[si=j ∣ si−1=i],1≤i,j≤N. For the special case of a discrete, first order, Markovchain, the probabilistic description for the current state (at time tt) and the predecessor state is s_{t}st. Furthermore we only consider those processes being independent of time, thereby leading to the set of state transition probability a_{ij}aij of the form: with the properties a_{ij} geq 0aij≥0 and sum_{i=1}^{N} A_{ij} = 1∑i=1NAij=1. The stochastic process can be called an observable Markovmodel. Now, let us consider the problem of a simple 4-state Markov model of weather. We assume that once a day (e.g., at noon), the weather is observed as being one of the following:$

State

The matrix A of state transition probabilities is:

${a_{ij}}= egin{Bmatrix} a_{11}&a_{12}&a_{13}&a_{14} \ a_{21}&a_{22}&a_{23}&a_{24} \ a_{31}&a_{32}&a_{33}&a_{34} \ a_{41}&a_{42}&a_{43}&a_{44} end{Bmatrix}A={aij}=⎩⎪⎪⎨⎪⎪⎧a11a21a31a41a12a22a32a42a13a23a33a43a14a24a34a44⎭⎪⎪⎬⎪⎪⎫$

Given the model, several interestingquestions about weather patterns over time can be asked (and answered). We canask the question: what is the probability (according to the given model) thatthe weather for the next kdays willbe? Another interesting question we can ask: given that the model is in a knownstate, what is the expected number of consecutive days to stay in that state?Let us define the observation sequence $s_{1}, s_{2}, s_{3}, ... , s_{k} ight }O={s1,s2,s3,...,sk}, and the probability of the observation sequence OO given the model is defined as p(O|model)p(O∣model). Also, let the expected number of consecutive days to stayin state ii be E_{i}Ei. Assume that the initial state probabilities p[s_{1} = i] = 1, 1 leq i leq Np[s1=i]=1,1≤i≤N. Bothp(O|model)p(O∣model) and E_{i}Ei are real numbers.$

Input Format

Line $O_{1}O1, and line 66 for the observation sequence O_{2}O2. Line 77and line 88 for the states of interest to find the expected number of consecutive days to stay in these states.$

Line $a_{11} a_{12} a_{13} a_{14}a11 a12 a13 a14$

Line $a_{21} a_{22} a_{23} a_{34}a21 a22 a23 a34$

Line $a_{31} a_{32} a_{33} a_{34}a31 a32 a33 a34$

Line $a_{41} a_{42} a_{43} a_{44}a41 a42 a43 a44$

Line $s_{1} s_{2} s_{3} ... s_{k}s1 s2 s3 ... sk$

Line $s_{1} s_{2} s_{3} ... s_{l}s1 s2 s3 ... sl$

Line

Output Format

Line $O_{1}O1and O_{2}O2 respectively. Line 33 and line 44 are for the expected number of consecutive days to stay in states ii and jj respectively.$

Line $p[O_{1} | model]p[O1∣model]$

Line $p[O_{2} | model]p[O2∣model]$

Line $E_{i}Ei$

Line $E_{j}Ej$

Please be reminded that the floating number should accurate to $10^{-8}10−8.$

样例输入

0.4 0.3 0.2 0.1
0.3 0.3 0.3 0.1
0.1 0.1 0.6 0.2
0.1 0.2 0.2 0.5
4 4 3 2 2 1 1 3 3
2 1 1 1 3 3 4
3
4

样例输出

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个是个题意题啊，看懂就出来了，我是读了n年的样例

#include<stdio.h>
#include<string.h>
int main()
{
    double a[5][5],s;
    int i,j,x,y,b[1001],l;
    char c[10000];
    for(i=1;i<=4;i++)
    {
        for(j=1;j<=4;j++)
        {
            scanf("%lf",&a[i][j]);
        }
    }
    getchar();
    gets(c);
    l=strlen(c);
    i=0;j=0;
    while(i<l)
    {
        if(c[i]>='0'&&c[i]<='9')
        {
            x=0;
            while(i<l&&c[i]>='0'&&c[i]<='9')
            {
                x=x*10+c[i]-'0';
                i++;
            }
            b[j++]=x;
        }
        i++;
    }
    s=1;
    for(i=1;i<j;i++)
    {
        s*=a[b[i-1]][b[i]];
    }
    printf("%.8lf
",s);
    c[0]=0;
    //getchar();
    gets(c);
    l=strlen(c);
    i=0;j=0;
    while(i<l)
    {
        if(c[i]>='0'&&c[i]<='9')
        {
            x=0;
            while(i<l&&c[i]>='0'&&c[i]<='9')
            {
                x=x*10+c[i]-'0';
                i++;
            }
            b[j++]=x;
        }
        i++;
    }
    s=1;
    for(i=1;i<j;i++)
    {
        s*=a[b[i-1]][b[i]];
    }
    printf("%.8lf
",s);
    scanf("%d",&x);
    printf("%.8lf
",1.0/(1-a[x][x]));
    scanf("%d",&x);
    printf("%.8lf
",1.0/(1-a[x][x]));
}

问答

28.14%
1000ms
131072K

Company ABC is holding an antique auction. In the auction, there are

• Each lot has a reserved price, which is a positive integer.

• Each bidder may submit at most one bid to each lot whose amount must be a positive integer.

• A valid bid for a lot is one that is at least the reserved price for the lot. All invalid bids are not considered.

• The largest valid bid wins the lot and is called the winning bid. In case of tie bids, the bidder with the smaller bidder number wins. If there is no valid bid for a lot, then this lot is not sold.

• The second largest bid is a valid bid that does not win if one exists; otherwise, it is the reserved price.

• If a bidder wins a lot, then the final hammer price for this lot is either $10\%10% over the second largest bid or the largest valid bid, depending on which one is smaller. If the final hammer price is not an integer, then it is truncated to the nearest integer.$

• Given the reserved price and the bids for each lot, your task is to decide the total final hammer prices for a given k,

Input Format

The input contains

Line

Line $r_{1}, b_{1,1}, p_{1,1}, b_{1,2}, p{1,2}, -1r1,b1,1,p1,1,b1,2,p1,2,−1$

Line $r_{i}, b_{i,1}, p_{i,1}, b_{i,2}, p{i,2}, -1ri,bi,1,pi,1,bi,2,pi,2,−1$

Line $r_{N}, b_{N,1}, p_{N,1}, b_{N,2}, p_{N,2}, -1rN,bN,1,pN,1,bN,2,pN,2,−1$

Line

Line $q_{1}q1$

Line $q_{i}qi$

Line $q_{k}qk$

In Line $r_{i}ri is the reserved price for lot ii. The number b_{i,j}bi,j and p{i,j}pi,j are the j^{th}jth bid for lot ii from bidder b_{i,j}bi,j with the amount p_{i,j}pi,j . Note that a space is between each number. The number -1−1 is added to mark the end of bids for a lot. In Line N+3N+3, kk is given. In line N + 3 + iN+3+i, you are asked to provide the total hammer prices for bidder q_{i}qi.$

Output Format:

The output contains

Line $h_{1}h1$

Line $h_{i}hi$

Line $h_{k}hk$

The total hammer prices for bidder $q_{i}qi is h_{i}hi.$

Hint

In the sample, the bidder

样例输入

3
3
11 2 12 1 15 -1
5 3 4 -1
23 1 32 2 35 3 40 -1
1
1

样例输出

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个C题意倒也好懂，主要还是精度问题

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#define ll long long
#define INF 0x3f3f3f3f
#define N 200005
using namespace std;

vector<pair<int,int> > vec;
int n,m;
int low[N];
ll a[N];

int main()
{
    scanf("%d%d",&n,&m);
    {
        memset(a,0,sizeof a);
        pair<int,int> pp;
        for(int i=1;i<=n;i++)
        {
            vec.clear();
            scanf("%d",&low[i]);
            int x,f=0;
            while(scanf("%d",&x),x!=-1)
            {
                int d;
                scanf("%d",&d);
                if(d>=low[i])
                {
                    pp.first=-d;
                    pp.second=x;
                    vec.push_back(make_pair(-d,x));
                }
            }
            sort(vec.begin(),vec.end());
            if(vec.size()>=2)
            {
                if(-vec[1].first*1.1> -vec[0].first) a[vec[0].second]+= -vec[0].first;
                else a[vec[0].second]+=(int)(-vec[1].first*1.1);
            }else if(vec.size()==1){
                if(low[i]*1.1> -vec[0].first) a[vec[0].second]+= -vec[0].first;
                else a[vec[0].second]+=(int)(low[i]*1.1);
            }
        }
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int s;
            scanf("%d",&s);
            printf("%lld
",a[s]);
        }
    }
    return 0;
}

问答

31.66%
1000ms
131072K

Three circles $C_{a}Ca, C_{b}Cb, and C_{c}Cc, all with radius RRand tangent to each other, are located in two-dimensional space as shown in Figure 11. A smaller circle C_{1}C1 with radius R_{1}R1 (R_{1}<RR1<R) is then inserted into the blank area bounded by C_{a}Ca, C_{b}Cb, and C_{c}Cc so that C_{1}C1 is tangent to the three outer circles, C_{a}Ca, C_{b}Cb, and C_{c}Cc. Now, we keep inserting a number of smaller and smaller circles C_{k} (2 leq k leq N)Ck (2≤k≤N) with the corresponding radius R_{k}Rk into the blank area bounded by C_{a}Ca, C_{c}Cc and C_{k-1}Ck−1 (2 leq k leq N)(2≤k≤N), so that every time when the insertion occurs, the inserted circle C_{k}Ck is always tangent to the three outer circles C_{a}Ca, C_{c}Cc and C_{k-1}Ck−1, as shown in Figure 11$

Figure 1.

(Left) Inserting a smaller circle $C_{1}C1 into a blank area bounded by the circle C_{a}Ca, C_{b}Cb and C_{c}Cc.$

(Right) An enlarged view of inserting a smaller and smaller circle $C_{k}Ck into a blank area bounded by C_{a}Ca, C_{c}Cc and C_{k-1}Ck−1 (2 leq k leq N2≤k≤N), so that the inserted circle C_{k}Ck is always tangent to the three outer circles, C_{a}Ca, C_{c}Cc, and C_{k-1}Ck−1.$

Now, given the parameters $R_{k}Rk, i.e., the radius of the k-thk−th inserted circle. Please note that since the value of R_kRk may not be an integer, you only need to report the integer part of R_{k}Rk. For example, if you find that R_{k}Rk = 1259.89981259.8998 for some kk, then the answer you should report is 12591259.$

Another example, if $R_{k}Rk = 39.102939.1029 for some kk, then the answer you should report is 3939.$

Assume that the total number of the inserted circles is no more than

$10^{4} leq R leq 10^{7}104≤R≤107.$

Input Format

Contains

Line

Output Format

Contains

Line $R_{k}Rk ----------------output for the value of kk and R_{k}Rk at the test case #11, each of which should be separated by a blank.$

Line $R_{k}Rk ----------------output for kk and the value of R_{k}Rk at the test case # ii, each of which should be separated by a blank.$

Line $R_{k}Rk ----------------output for kk and the value ofR_{k}Rk at the test case # ll, each of which should be separated by a blank.$

样例输入

样例输出

1 23665

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

笛卡尔定理搞一搞就行啊

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int main()
{
    int n,i,j,u;
    double a,R,r,b[11],co,p,x;
    while(scanf("%d",&n),n!=-1)
    {
        scanf("%lf",&a);
        R=a,r=a;
        for(i=1;i<=10;i++)
        {
            co=(2*(R+r)*(R+r)-(R+R)*(R+R))/2/(R+r)/(R+r);
            p=acos(co)/2;
            co=cos(p);
            x=(2*r-2*R)-2*(R+r)*co;
            x=(2*(R+r)*co*r-2*r*r-2*R*r)/x;
            b[i]=x;
            r=x;
        }
        while(n--)
        {
            scanf("%d",&u);
            printf("%d %d
",u,int(b[u]));
        }
    }
}

问答

52.07%
1000ms
131072K

In the Personal Communication Service systems such as GSM (Global System for Mobile Communications), there are typically a number of base stations spreading around the service area. The base stations are arranged in a cellular structure, as shown in the following figure. In each cell, the base station is located at the center of the cell.

For convenience, each cell is denoted by [

Now the question is as follows. We have a service area represented by a Euclidean plane (i.e.,

You are given an input (

For example, given a location (

input = (
output = [

Input Format

A list of

Output Format

A list of

Please be reminded that there exist a space between coordinates.

样例输入

样例输出

[0,0], [-1,2], [0,0], [1,1], [0,1], [0,2], [2,2], [-1,-1], [2,-1], [-1,0]

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

这个I我并不懂，也不知道他们怎么过的，之后补一下吧

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define N 1005

using namespace std;
const double G = sqrt(3.0);

int xx[100000];

map<pair<int,int> , pair<int,int> > ans;
map<pair<int,int> ,bool> used;

void addd()
{
    xx[0]=1;
    xx[1]=1;
    for(int i=2;i<=10000;i++)
    {
        xx[i]=xx[i-1]+xx[i-2];
    }
}


void check(int x,int y){
    
    int ii;
    int aa=0;
    for(ii=0;ii<100;ii++) aa++;
    
    double a = 5*G/2.0*y+5*G*x;
    double b = 5*1.5*y;
    int imin = a-6;
    int imax = a+6;
    int jmin = b-6;
    int jmax = b+6;
    for (int i=imin;i<=imax;i++){
        for (int j=jmin;j<=jmax;j++){
            bool flag = true;
            if (i>=a-5*G/2&&i<=a+5*G/2);else flag = false;
            if (!flag)continue;
            double top = G/3*i+(b-G/3*a)+5;
            double bottom = G/3*i+(b-G/3*a)-5;
            if (j>=bottom&&j<=top);else flag = false;
            if (!flag)continue;
            top = -G/3*i+(b+5+G/3*a);
            bottom = -G/3*i+(b-5+G/3*a);
            if (j>=bottom&&j<=top);else flag = false;
            if (!flag)continue;
            if (!used[make_pair(i,j)]){
                ans[make_pair(i,j)] = make_pair(x,y);
                used[make_pair(i,j)] = true;    
            }
        }
    }
}
int main(){
    
    
    for (int i=-20;i<=20;i++){
        for (int j=-20;j<=20;j++){
            check(i,j);
        }
    }
    int aa=0;
    for(int ii=0;ii<10000;ii++) aa++;
    
    for(int i=0;i<=500;i++)addd();
    
    
    
    
    int x,y;
    int T=10;
    while (scanf("%d%d",&x,&y)!=EOF){
        T--;
        pair<int,int> res = ans[make_pair(x,y)];
        printf("[%d,%d]",res.first,res.second);
        if (T!=0){
            printf(", ");
        }
    }
    printf("
");
    return 0;
}

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原文地址：https://www.cnblogs.com/BobHuang/p/7591375.html