2013ACM多校联合（2）

A:小Y做家教

简单的想法题。使用hash表或者是map存储所有数，然后从最小的数开始找从这个数开始的连续P倍数的个数X，那么需要删除的数的个数为X/2。

代码如下：

 

B:小Y的难题Ⅰ

组合数学题，通过经典的一一对应原则推导出答案为N^(N-2)。http://wenku.baidu.com/view/2e1ab2757fd5360cba1adbba.html
Cayley定理在组合数学中的应用。

代码如下：

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <string>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const LL M=1e9+7;
LL Pow( LL a, LL b)
{
    LL ans=1;
    while( b ){
        if( b&1 ){
            ans*=a;
            ans%=M;
        }
        a*=a;
        a%=M;
        b>>=1; 
    }
    return ans;
} 
int main( )
{
    LL N;
    while( scanf( "%lld", &N )!=EOF ){
        printf( "%lld\n", Pow( N, N-2 ) );
    }
    return 0;
}

 

C:小Y的逆袭

计算几何。该题一个简单的做法是先找出三个点的外心，然后用极坐标每次旋转72°生成剩下的点。

代码如下：

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;

const double PI = acos(-1);

struct point{double x,y;};
struct line{point a,b;};
vector<point>v;

point p[3], cc;

point intersection(line u,line v){
    point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))
        /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}

point circumcenter(point a,point b,point c){
    line u,v;
    u.a.x=(a.x+b.x)/2;
    u.a.y=(a.y+b.y)/2;
    u.b.x=u.a.x-a.y+b.y;
    u.b.y=u.a.y+a.x-b.x;
    v.a.x=(a.x+c.x)/2;
    v.a.y=(a.y+c.y)/2;
    v.b.x=v.a.x-a.y+c.y;
    v.b.y=v.a.y+a.x-c.x;
    return intersection(u,v);
}

double dist(point a, point b) {
    return sqrt(double( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) )); 
}

bool cmp(point a, point b) {
    if (fabs(a.x - b.x) > 1e-6) {
        return a.x < b.x;
    } else {
        return a.y < b.y;
    }
}

int main() {
    double R;
    point info;
    while (scanf("%lf %lf", &p[0].x, &p[0].y) != EOF) {
        v.clear();
        for (int i = 1; i < 3; ++i) {
            scanf("%lf %lf", &p[i].x, &p[i].y);
        }
        cc = circumcenter(p[0], p[1], p[2]);
        R = dist(p[0], cc);
        double phi = atan2(p[0].y-cc.y, p[0].x-cc.x); // 求出第一个点的角度
        for (int i = 0; i < 5; ++i) {
            info.x = R * cos(phi) + cc.x, info.y = R * sin(phi) + cc.y;
            v.push_back(info);
            phi += 72.0*PI/180.0;
        }
        sort(v.begin(), v.end(), cmp);
        for (int i = 0; i < 5; ++i) {
            int flag = 0;
            for (int j = 0; j < 3; ++j) {
                if (fabs(v[i].x - p[j].x) < 1e-6 && fabs(v[i].y - p[j].y) < 1e-6) {
                    flag = 1;
                    break;
                }
            }
            if (!flag) {
                printf("%.2f %.2f\n", v[i].x, v[i].y);
            }
        }
    }
    return 0;    
}

 

D:Palindrome

动态规划。题意是求将一个串改造成回文串的最少花费。给出所有组成串的字母添加和删除的花费。设dp[i][j]表示[i,j]子串被改造成回文串的最少开销是多少。那么有动态规划方程：
dp[i][j] = min(dp[i+1][j] + cost[i], dp[i][j-1] + cost[j]); 其中cost[i] = min(add[i], del[i])存储删除和添加节点的较小的花费。如果str[i] == str[j]的话就多出一种选择dp[i-1][j-1]。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
int N, M, dp[2005][2005];
int cost[30];
char str[2005];

int DP() {
    for (int i = 1; i < M; ++i) {
        for (int j = 1; j+i <= M; ++j) {
            int a = j, b = j + i, ll = str[a]-'a', rr = str[b]-'a';
            dp[a][b] = min(dp[a+1][b] + cost[ll], dp[a][b-1] + cost[rr]);
            if (ll == rr)
                dp[a][b] = min(dp[a][b], dp[a+1][b-1]);
        }
    }
    return dp[1][M];
}

int main() {
    char s[5];
    int a, b;
    while (scanf("%d %d", &N, &M) == 2) {
        scanf("%s", str+1);
        for (int i = 1; i <= N; ++i) {
            scanf("%s %d %d", s, &a, &b);
            cost[s[0]-'a'] = min(a, b);
        }
        printf("%d\n", DP());
    }
    return 0;
}

 

E:小Y的难题Ⅱ

分解素因子，转换成M个球放到N个盒子模型，得出组合公式C(N+M-1, N-1),因为ai>1,既盒子不允许为空，所以求出总方案数后，减去盒子为空的情形。对于盒子为空的求解，通过假设至少一个盒子（C(N+M-1-1,M-1-1)），两个，直到N-1个，由容斥来解决重复问题。注意单个素因子，其次数可能会很大，所以C预处理要比较大，不然会RE。

代码如下：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<assert.h>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
typedef long long LL;

const int mod = 1e9+7;

int b[21], a[1010], n, tot;
LL C[1100][1100];

map<int,int>mp;

void init(){
    for(int i = 0; i <= 1000; i++)
        C[i][0] = 1, C[i][i] = 1;
    for(int i = 2; i <= 1000; i++)
        for(int j = 1; j < i; j++)
            C[i][j] = (C[i-1][j-1]+C[i-1][j])%mod;
}

void deal(){
    mp.clear();
    for(int i = 0; i < n; i++){
        
        int tmp_max = (int)sqrt(1.*b[i]);
        int t = b[i];
        for(int j = 2; j <= tmp_max; j++)
        {
            if( t%j == 0 ){
                if( mp.count(j) == 0 ) mp[j] = 0;

                int cnt = 0;    
                while( t%j == 0 ) (t/=j), cnt++;
                mp[j] += cnt;
            } 
        }
        if( t > 1 ){
            if( mp.count(t) == 0 ) mp[t] = 1;
            else    mp[t]++;
        }    
    }
    int idx = 0;
    memset( a, 0, sizeof(a));
    for( map<int,int>::iterator it = mp.begin(); it != mp.end(); it++ )
        a[idx++] = it->second;
    tot = mp.size();
}
void solve(){
    LL ans = 0;
        
    for(int i = 0; i < n; i++){
        LL tmp = 1;
        for(int j = 0; j < tot; j++){    
            assert( (((n-1-i+a[j])>=0)&&(n-1-i+a[j])<=1000) );    
            tmp = tmp*C[ n-1-i+a[j] ][ n-1-i ]%mod;    
        }
        tmp = tmp*C[n][i]%mod;
        
        if( i&1 ) tmp = -tmp;
        ans = (ans+tmp+mod)%mod;
    }
    printf("%lld\n", ans);
}
int main(){
    init();
    int T;
    scanf("%d", &T);
    while( T-- ){
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &b[i] );
        deal();
        solve();
    }
    return 0;
}

 

F:War Ⅰ

并查集+最短路。通过并查集处理结盟的点，然后从集合中选择一个到目标点最短的距离输出。

代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>

using namespace std;

const int  inf = 0x7f7f7f7f;
const int MAXV = 1002;

int UFS[MAXV];

int Find(int x)
{
    return UFS[x]=(UFS[x]!=x?Find(UFS[x]):x);
}
void Union(int x,int y)
{
    UFS[Find(x)]=Find(y);
}

char o;
int n,m,k,u,v,w,x,y,a[MAXV][MAXV];

int main()
{
    while(scanf("%d %d %d",&n,&m,&k)!=EOF)
    {
        memset(a,inf,sizeof(a));
        for(int i=0;n>i;i++)
        {
            UFS[i]=i;
        }
        for(int i=0;m>i;i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            a[u][v]=min(a[u][v],w);
        }
        for(int z=0;n>z;z++)
        {
            for(int i=0;n>i;i++)
            {
                if(a[i][z]!=inf)
                {
                    for(int j=0;n>j;j++)
                    {
                        if(a[z][j]!=inf)
                        {
                            a[i][j]=min(a[i][j],a[i][z]+a[z][j]);
                        }
                    }
                }
            }
        }
        for(int i=0;k>i;i++)
        {
            getchar();
            scanf("%c %d %d",&o,&x,&y);
            if(o=='I') Union(x,y);
            else
            {
                int ans=inf;
                for(int i=0;n>i;i++)
                {
                    if(Find(x)==Find(i))
                    {
                        ans=min(ans,a[i][y]);
                    }
                }
                printf("%d\n",ans!=inf?ans:-1);
            }
        }
    }
    return 0;
}

 

G:War Ⅱ

二分枚举构图+二分匹配。该题想说明的就是A中的一个城市只能够对应B中的一个城市。先通过floyd求出任意城市之间的最短路，然后二分枚举构好A->B城市的二分图，然后使用匈牙利算法计算是否能够达到完全匹配。

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <cassert>
using namespace std;

const int MAX = 105;
const int INF = 0x3f3f3f3f;
int n, N, M, reca[MAX], recb[MAX];
int mp[MAX][MAX];
int marry[MAX];
char G[MAX][MAX];
char vis[MAX];

void floyd() {
    for (int k = 0; k < n; ++k) {
        for (int i = 0; i < n; ++i) {
            if (mp[i][k] == INF || i == k) continue;
            for (int j = 0; j < n; ++j) {
                if (mp[k][j] == INF || j == k) continue;
                if (mp[i][j] > mp[i][k] + mp[k][j]) {
                    mp[i][j] = mp[i][k] + mp[k][j];
                }
            }
        }    
    }
}

bool path(int u) 
{
    for (int i = 0; i < M; ++i) {
        if (!G[u][i] || vis[i])
            continue;
        vis[i] = 1;
        if (marry[i] == -1 || path(marry[i])) {
            marry[i] = u;
            return true;
        }
    }
    return false;
}

void build(int lim) {
    memset(G, 0, sizeof (G));
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < M; ++j) {
            if (mp[reca[i]][recb[j]] <= lim) {
                G[i][j] = 1;
            }
        }
    }
}

bool Ac() {
    int cnt = 0;
    memset(marry, 0xff, sizeof (marry));
    for (int i = 0; i < N; ++i) {
        memset(vis, 0, sizeof (vis));
        if (path(i)) {
            ++cnt;
        }
    }
    return cnt == M;
}

int bsearch(int l, int r) {
    int mid, ret;
    while (l <= r) {
        mid = (l + r) >> 1;
        build(mid);
        if (Ac()) {
            ret = mid;
            r = mid - 1;    
        } else {
            l = mid + 1;
        }
    }
    return ret;
}

int main()
{
    while (scanf("%d", &n) == 1) {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                scanf("%d", &mp[i][j]);
            }
        }
        scanf("%d", &N);
        for (int i = 0; i < N; ++i) {
            scanf("%d", &reca[i]);
            --reca[i];
        }
        scanf("%d", &M);
        for (int i = 0; i < M; ++i) {
            scanf("%d", &recb[i]);
            --recb[i];
        }
        floyd();
        printf("%d\n", bsearch(0, 10000));
    }
    return 0;
}

 

H:War Ⅲ

法一：通过题目给定的关系不难知道a[i] = 3*a[i-1] + a[i-2] - 3*a[i-3]。假设要求第D天的口令，那么计算出a[D-1], a[D-2], a[D-3]就能够计算出a[D]，求出四个系数后，代入X便可求解答案了。这题D给的数据比较大，直接计算肯定会TLE，那么使用矩阵并加上快速降幂来计算就简单多了，将给定的b,c,d作为3*1的矩阵B{a[0], a[-1], a[-2]}^-1，矩阵A=
3 1 -3
1 0 0
0 1 0
作为递推矩阵，那么在B矩阵左边乘上D-1个A最后的结果就是{a[D-1], a[D-2], a[D-3]}^-1，D-1个A相乘就可以使用矩阵快速幂了。

代码如下：

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cassert> 
using namespace std;

typedef long long int64;
const int MOD = int(1e9)+7;
int a, b, c, d, D, X;

int64 R[3][3] = {
    {3, 1, -3},
    {1, 0, 0},
    {0, 1, 0}
};

struct Matrix {
    int64 mat[3][3];
    Matrix() {
        memset(mat, 0, sizeof (mat));
    }
    void unit() {
        memset(mat, 0, sizeof (mat));
        for (int i = 0; i < 3; ++i) {
            mat[i][i] = 1;
        }
    }
    void init() {
        memcpy(mat, R, sizeof (mat));
    }
    void show() const;
    friend Matrix operator * (const Matrix & a, const Matrix & b);
    friend Matrix pow(Matrix a, int b);
};

void Matrix::show() const {
    for (int i = 0; i < 3; ++i) {
        for (int j = 0; j < 3; ++j) {
            printf("%d ", mat[i][j]);    
        }
        puts("");
    }
}

Matrix operator * (const Matrix & a, const Matrix & b) {
    Matrix ret;
    for (int i = 0; i < 3; ++i) {
        for (int j = 0; j < 3; ++j) {
            for (int k = 0; k < 3; ++k) {
                ret.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % MOD;
                ret.mat[i][j] %= MOD;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix a, int b) {
    Matrix ret;
    ret.unit();
    while (b) {
        if (b & 1) {
            ret = ret * a;
        }
        a = a * a;
        b >>= 1;
    }
    return ret;
}

int64 cal(int64 na, int64 nb, int64 nc, int64 nd) {
    int64 ini[4] = {1};
    for (int i = 1; i < 4; ++i) {
        ini[i] = (ini[i-1] * X) % MOD;
    }
    return ((na*ini[3])%MOD+(nb*ini[2])%MOD+(nc*ini[1])%MOD+(nd*ini[0])%MOD)%MOD;
}

void solve() {
    Matrix mm; 
    mm.init();
    mm = pow(mm, D-1);
    int64 na, nb, nc, nd;
    nb = ((mm.mat[0][0]*b)%MOD + (mm.mat[0][1]*c)%MOD + (mm.mat[0][2]*d)%MOD)%MOD;
    nc = ((mm.mat[1][0]*b)%MOD + (mm.mat[1][1]*c)%MOD + (mm.mat[1][2]*d)%MOD)%MOD;
    nd = ((mm.mat[2][0]*b)%MOD + (mm.mat[2][1]*c)%MOD + (mm.mat[2][2]*d)%MOD)%MOD;
    na = (3*nb)%MOD+nc-(3*nd)%MOD;
    printf("%d\n", int((cal(na, nb, nc, nd)+MOD)%MOD));
}

int main() {
    while (scanf("%d %d %d %d", &a, &b, &c, &d) != EOF) {
        assert(a >= 0 && a <= 100);
        scanf("%d %d", &D, &X);
        solve();
    }
    return 0;
} 

 

法二：根据表达式a[i] = 3*a[i-1] + a[i-2] - 3*a[i-3]，由于该表达式为常系数3阶齐次递推关系，因此设a[n] = r^n，代入方程求出特征根为-1,1,3，因此a[n]=A*(-1)^n+B*(1)^n+C*(3)^n，令a[0]=d,a[1]=c,a[2]=b求出A,B,C之后再根据通项公式求出D天后的系数。b,c,d是8的倍数保证了解三元一次方程组时不会出现分数。

代码如下：

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

typedef long long int64;
const int64 MOD = (int64)1e9 + 7;
int64 a, b, c, d, D, X;

int64 _pow(int64 a, int64 b) {
    int64 ret = 1;
    while (b) {
        if (b & 1) {
            ret *= a;
            ret %= MOD;
        }
        b >>= 1;
        a *= a;
        a %= MOD;
    }
    return ret;
}

int64 cal(int64 na, int64 nb, int64 nc, int64 nd) {
    int64 ini[4] = {1};
    for (int i = 1; i < 4; ++i) {
        ini[i] = (ini[i-1] * X) % MOD;
    }
    return ((na*ini[3])%MOD+(nb*ini[2])%MOD+(nc*ini[1])%MOD+(nd*ini[0])%MOD)%MOD;
}

int main() {
    while (scanf("%lld %lld %lld %lld", &a, &b, &c, &d) != EOF) {
        scanf("%lld %lld", &D, &X);
        int64 A = (3*d-4*c+b)/8;
        int64 B = (3*d+2*c-b)/4;
        int64 C = (b-d)/8;
        a = (A * ((D+2) & 1 ? -1 : 1) + B + C * _pow(3, D+2)) % MOD;
        b = (A * ((D+1) & 1 ? -1 : 1) + B + C * _pow(3, D+1)) % MOD;
        c = (A * ((D) & 1 ? -1 : 1) + B + C * _pow(3, D)) % MOD;
        d = (A * ((D-1) & 1 ? -1 : 1) + B + C * _pow(3, D-1)) % MOD;
        printf("%lld\n", (cal(a, b, c, d)+MOD)%MOD);
    }
    return 0;    
}