ACM-ICPC 2018 沈阳赛区(网络赛)

D.Made In Heaven

One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are NN spots in the jail and MM roads connecting some of the spots. JOJO finds that Pucci knows the route of the former (K-1)(K−1)-th shortest path. If Pucci spots JOJO in one of theseK−1 routes, Pucci will use his stand Whitesnake and put the disk into JOJO's body, which means JOJO won't be able to make it to the destination. So, JOJO needs to take the KK-th quickest path to get to the destination. What's more, JOJO only has TT units of time, so she needs to hurry.

JOJO starts from spot SS, and the destination is numbered E. It is possible that JOJO's path contains any spot more than one time. Please tell JOJO whether she can make arrive at the destination using no more than TT units of time.

Input

There are at most 50 test cases.

The first line contains two integers N and M(1≤N≤1000,0≤M≤10000). Stations are numbered from 1 to N.

The second line contains four numbers S, E, K and T (1≤K≤10000, 1≤T≤100000000 ).

Then M lines follows, each line containing three numbers U,V and W (1≤U,V≤N,1≤W≤1000) . It shows that there is a directed road from UU-th spot to V-th spot with time W.

It is guaranteed that for any two spots there will be only one directed road from spot A to spot B (1≤A,B≤N,A≠B), but it is possible that both directed road <A,B> and directed road <B,A>exist.

All the test cases are generated randomly.

Output

One line containing a sentence. If it is possible for JOJO to arrive at the destination in time, output "yareyaredawa" (without quote), else output "Whitesnake!" (without quote).

样例输入

样例输出

yareyaredawa

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题目思路：求个第k短路就好了

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x7FFFFFFF;

const int MAXN = ;

const int MAXM = ;

int S, E, K, T;

struct Edge {

    int u, v, c, next, next1;

    Edge() {}

    Edge(int u, int v,int c):u(u), v(v), c(c) {}

} edge[MAXM * ];

int first[MAXN], first1[MAXN], sign;

int n, m;

int dis[MAXN];

bool vis[MAXN];

struct Node {

    int v, c;

    Node() {}

    Node(int v, int c):v(v), c(c) {}

    bool operator < (const Node& a) const {

        return c + dis[v] > a.c + dis[a.v];

    }

};

void init() {

    memset(first, -, sizeof(first));

    memset(first1, -, sizeof(first1));

    sign = ;

}

void addEdge(int u, int v, int c) {

    edge[sign] = Edge(u, v, c);

    edge[sign].next1 = first1[v];

    first1[v] = sign;

    edge[sign].next = first[u];

    first[u] = sign++;

}

priority_queue<Node>que;

void dijkstra(int s) {

    memset(vis, , sizeof(vis));

    for(int i = ; i <= n; i++) {

        dis[i] = INF;

    }

    while(!que.empty()) { que.pop(); }

    dis[s] = ;

    que.push(Node(s, ));

    while(!que.empty()) {

        Node cur = que.top();

        que.pop();

        if(!vis[cur.v]) {

            vis[cur.v] = ;

            dis[cur.v] = cur.c;

            for(int i = first1[cur.v]; ~i; i = edge[i].next1) {

                if(dis[edge[i].u] > dis[cur.v] + edge[i].c) {

                    que.push(Node(edge[i].u, edge[i].c + cur.c));

                }

            }

        }

    }

}

int a_star(int src) {

//    priority_queue<Node>que;

    while(!que.empty()) { que.pop(); }

    que.push(Node(src, ));

    while(!que.empty()) {

        Node cur = que.top();

        que.pop();

        if(cur.v == E) {

            if(K > ) {

                K--;

            }

            else {

                return cur.c;

            }

        }

        for(int i = first[cur.v]; ~i; i = edge[i].next) {

            que.push(Node(edge[i].v, cur.c + edge[i].c));

        }

    }

    return -;

}

int main() {

    int u, v, c;

    while(~scanf("%d%d",&n, &m)) {

        init();

        scanf("%d %d %d %d", &S, &E, &K, &T);

        while(m--) {

            scanf("%d %d %d",&u, &v, &c);

            addEdge(u, v, c);

        }

        dijkstra(E);

        if(dis[S] == INF) {

            puts("Whitesnake!");

            continue;

        }

        if(S == E) {

            K++;

        }

        int res = a_star(S);

        if(res == -) {

            puts("Whitesnake!");

            continue;

        }

        if(res <= T) {

            puts("yareyaredawa");

        } else {

            puts("Whitesnake!");

        }

    }

    return ;

}

"Oh, There is a bipartite graph.""Make it Fantastic."

X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up several edges from the current graph and try to make the degrees of every point to between the two boundaries. If you pick one edge, the degrees of two end points will both increase by one. Can you help X to check whether it is possible to fix the graph?

Input

There are at most 30 test cases.

For each test case,The first line contains three integers NN the number of left part graph vertices, MM the number of right part graph vertices, and KK the number of edges (1≤N≤2000,0≤M≤2000,0≤K≤6000 ). Vertices are numbered from 11 to NN.

The second line contains two numbers L, R (0≤L≤R≤300). The two fantastic numbers.

Then KK lines follows, each line containing two numbers U, V(1≤U≤N,1≤V≤M). It shows that there is a directed edge from U-th spot to V-th spot.

Note. There may be multiple edges between two vertices.

Output

One line containing a sentence. Begin with the case number. If it is possible to pick some edges to make the graph fantastic, output "Yes" (without quote), else output "No" (without quote).

样例输入

样例输出

Case 1: Yes

Case 2: No

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题目大意：给出一个二分图，再给出一些边，再给出L，R，从这些边中选取一些使得每一个点的度都在[L,R]之间。

题目思路：建立超级源点S，连接二分图左边的所有点，流量下界L,上界为R，同理将右半边所有点连接到超级汇点T。

将给出的边连接，流量下界0,上界为1。然后套个有上下界的网络流板子判断是否有解就行了。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int LEN = 1e5 + ;

const int oo = (1LL << ) - ;

namespace DINIC {

int tot, src, tar, qh, qt, cnt, s, t, S, T;

int ans, flow;

struct edge {

    int vet, next, len;

} E[LEN * ];

int dis[LEN], head[LEN], cur[LEN], q[LEN], IN[LEN];

void add(int u, int v, int c) {

    E[++tot] = (edge) {

        v, head[u], c

    };

    head[u] = tot;

}

void join(int u, int v, int c) {

    add(u, v, c);

    add(v, u, );

}

void init() {

    tot = -;

    for (int i = ; i <= cnt + ; i++)

        head[i] = -;

}

bool bfs() {

    memset(dis, , sizeof(dis));

    qh = qt = ;

    q[++qt] = src;

    dis[src] = ;

    while (qh < qt) {

        int u = q[++qh];

        for (int e = head[u]; e != -; e = E[e].next) {

            int v = E[e].vet;

            if (E[e].len && !dis[v]) {

                dis[v] = dis[u] + ;

                if (v == tar)

                    return ;

                q[++qt] = v;

            }

        }

    }

    return dis[tar];

}

int dfs(int u, int aug) {

    if (u == tar || !aug)

        return aug;

    int tmp = ;

    for (int &e = cur[u]; e != -; e = E[e].next) {

        int v = E[e].vet;

        if (dis[v] == dis[u] + ) {

            if (tmp = dfs(v, min(aug, E[e].len))) {

                E[e].len -= tmp;

                E[e ^ ].len += tmp;

                return tmp;

            }

        }

    }

    return ;

}

int maxflow(int s, int t) {

    src = s, tar = t;

    int res = , flow = ;

    while (bfs()) {

        for (int i = ; i <= cnt + ; i++)

            cur[i] = head[i];

        while (flow = dfs(src, oo))

            res += flow;

    }

    return res;

}

}

using namespace DINIC;

int N, M, K, L, R;

int main() {

    int cas = ;

    while(~scanf("%d %d %d", &N, &M, &K)) {

        scanf("%d %d", &L, &R);

        memset(IN, , sizeof(IN));

        ans = flow = ;

        cnt = N + M + ;

        s = N + M + ;

        t = N + M + ;

        S = ++cnt, T = ++cnt;

        init();

        for(int i = ; i <= K; i++ ) {

            int u, v, l = , r = ;

            scanf("%d %d", &u, &v);

            join(u, v + N, r - l);

            IN[v] += l, IN[u] -= l;

        }

        for(int i = ; i <= N; i++ ) {

            int u, v, l = L, r = R;

            u = s, v = i;

            join(u, v, r - l);

            IN[v] += l, IN[u] -= l;

        }

        for(int i = ; i <= M; i++ ) {

            int u, v, l = L, r = R;

            u = i + N, v = t;

            join(u, v, r - l);

            IN[v] += l, IN[u] -= l;

        }

        for (int i = ; i <= N + M + ; i++) {

            if (IN[i] < ) {

                join(i, T, -IN[i]);

            } else if (IN[i] > ) {

                flow += IN[i];

                join(S, i, IN[i]);

            }

        }

        join(t, s, oo);

        ans = maxflow(S, T);

        printf("Case %d: ", cas++);

        if (ans != flow)

            puts("No");

        else {

//            ans = maxflow(s, t);

//            printf("%lld\n", ans);

            puts("Yes");

        }

    }

    return ;

}

G.Spare Tire

A sequence of integer \lbrace a_n \rbrace{an} can be expressed as:

Now there are two integers nn and mm. I'm a pretty girl. I want to find all b1,b2,b3⋯bp that 1≤bi≤n and bi is relatively-prime with the integer m. And then calculate:

But I have no time to solve this problem because I am going to date my boyfriend soon. So can you help me?

Input

Input contains multiple test cases ( about 15000 ). Each case contains two integers nn and mm.1≤n,m≤10^8.

Output

For each test case, print the answer of my question(after mod 1,000,000,007).

Hint

In the all integers from 1 to 4, 1 and 3 is relatively-prime with the integer 4. So the answer is a1+a3=14.

样例输入

4 4

样例输出

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

typedef long long ll;

const int mod=1e9+;

int vis[],pri[],top;

int a[],h;

ll n,m,ans,sum;

ll p1,p2;

void init()

{

    top=;

    for(int i=; i<; i++)

    {

        if(!vis[i])

        {

            pri[++top]=i;

            for(int j=i+i; j<; j+=i)

                vis[j]=;

        }

    }

}

ll qpow(ll a,ll b)

{

    ll ans=;

    while(b)

    {

        if(b&)

            ans=ans*a%mod;

        a=a*a%mod;

        b>>=;

    }

    return ans;

}

ll getsum(int x)

{

    if(!x||x==)

        return ;

    int i,j,k;

    ll ans;

    k=n/x;

    ans=1LL*k*(k+)%mod*p1%mod*x%mod;

    ans+=(1LL*k*(k+)%mod*(*k+)%mod*p2%mod*x%mod*x%mod);

    ans=ans%mod;

    return ans;

}

void dfs(int i,int num,int x)

{

    if(x>n)

        return ;

    if(i==h)

    {

        if(num%)

        {

            sum+=getsum(x);

            sum=sum%mod;

        }

        else

            sum-=getsum(x);

        return ;

    }

    dfs(i+,num+,x*a[i]);

    dfs(i+,num,x);

}

int main()

{

    int i,j,k,t,cas;

    ll w;

    init();

    p1=qpow(,mod-);

    p2=qpow(,mod-);

    while(scanf("%lld%lld",&n,&m)!=EOF)

    {

        ans=1LL*n*(n+)%mod*p1%mod;

        ans+=(1LL*n*(n+)%mod*(*n+)%mod*p2%mod);

        ans=ans%mod;

        h=;

        w=m;

        for(i=; i<=top; i++)

        {

            if(w%pri[i]==)

            {

                a[h++]=pri[i];

                while(w%pri[i]==)

                    w=w/pri[i];

            }

        }

        if(w!=)

            a[h++]=w;

        sum=;

        dfs(,,);

        ans=(ans+mod-sum)%mod;

        printf("%lld\n",ans);

    }

    return ;

}

I.Lattice's basics in digital electronics

LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10^{-9}10−9 second. In the next 10^{-9}10−9 second, he built a data decoding device that decodes data encoded with his special binary coding rule to meaningful words.

His coding rule is called "prefix code", a type of code system (typically a variable-length code) distinguished by its possession of the "prefix property", which requires that there is no whole code word in the system that is a prefix (initial segment) of any other code word in the system. Note that his code is composed of only 0 and 1.

LATTICE's device only receives data that perfectly matches LATTICE's rules, in other words, people who send message to LATTICE will always obey his coding rule. However, in the process of receiving data, there are errors that cannot avoid, so LATTICE uses parity check to detect error bytes, after every 88-bit data there is 11 bit called parity bit, which should be '0' if there are odd number of '1's in the previous 88bits and should be '1' if there are even number of '1's. If the parity bit does not meet the fact, then the whole 99 bits (including the parity bit) should be considered as invalid data and ignored. Data without parity bit is also considered as invalid data. Parity bits will be deleted after the parity check.

For example, consider the given data "101010101010101010101010", it should be divided into 33parts:"101010101","010101010" and "101010". For the first part, there are 44 '1's in the first 88 bits, and parity bit is '1', so this part passed the check. For the second part, there are 44 '1's and parity bit is '0', so this part failed the check. For the third part, it has less than 99 bits so it contains no parity bit, so this part also failed the check. The data after parity check is "10101010", which is the first 88 bits of first part.

Data passed the parity check will go into a process that decodes LATTICE's code. The process is described in the following example: consider a situation that, "010" represents 'A' and "1011" represents 'B', if the data after parity check is "01010110101011010010", it can be divided into "010"+"1011"+"010"+"1011"+"010"+"010", which means "ABABAA" . LATTICE's device is so exquisite that it can decode all visible characters in the ASCII table .

LATTICE is famous for his Talk show, some reporters have sneaked into his mansion, they stole the data LATTICE to decode in hexadecimal, the coding rule consists of N pairs of corresponding relations from a bit string Si to an ASCII code Ci, and the message length MM, they want to peek his privacy so they come to you to write a program that decodes messages that LATTICE receives.

Input

The first line an integer T (T<35) represents the number of test cases.

Every test case starts with one line containing two integers, M (0<M≤100000), the number of original characters, and N (1≤N≤256), then N lines, every line contains an integer Ci, and a string Si(0<∣Si∣≤10), means that Si represents Ci, the ASCII code to a visible character and Si only contains '0'or '1' and there are no two numbers ii and jj that S_iSi is prefix of S_jSj.

Then one line contains data that is going to be received in hexadecimal. (0<∣data∣<200000).

Output

For each test case, output the decoded message in a new line, the length of the decoded message should be the same with the length of original characters, which means you can stop decoding having outputted M characters. Input guarantees that it will have no less than M valid characters and all given ASCII codes represent visible characters.

样例输入

2

15 9

32 0100

33 11

100 1011

101 0110

104 1010

108 00

111 100

114 0111

119 0101

A6Fd021171c562Fde1

8 3

49 0001

50 01001

51 011

14DB24722698

样例输出

hello world!!!!

12332132

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题目大意：

给出二进制和ASCLL码的对映规则，再给出一个16进制的字符串。将16进制的字符串转二进制，然后每九位校验一下（前八位中1的个数是奇数且第九位为1 或前八位中1的个数是偶数且第九位为0 则正确）

将正确的留下，再用题目给的规则转换成字符，只要前m个。暴力模拟下即可。

#include<cstdio>

#include<cstring>

using namespace std;

struct Trie

{

    int num;

    int next[];

} tree[];

int cnt = ;

char ttr[];

char ltr[];

bool a[];

int top = ;

void getBit()

{

    int len = strlen(ttr);

    top = ;

    for(int i=; i<len; i++)

    {

        switch (ttr[i])

        {

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case '':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'a':

        case 'A':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'b':

        case 'B':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'c':

        case 'C':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'd':

        case 'D':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'e':

        case 'E':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        case 'f':

        case 'F':

            ltr[top]='',ltr[top+]='',ltr[top+]='',ltr[top+]='',top+=;

            break;

        }

    }

    ltr[top]='\0';

}

void insert_tree(char str[],int num)

{

    int p = ;

    int len = strlen(str);

    for(int i=; i<len; i++)

    {

        if(tree[p].next[str[i]-''])

        {

            p = tree[p].next[str[i]-''];

        }

        else

        {

            tree[p].next[str[i]-''] = ++cnt;

            p = tree[p].next[str[i]-''];

        }

    }

    tree[p].num = num;

}

void solve()

{

    cnt = ;

    int tot = ;

    memset(tree,,sizeof(tree));

    memset(a,,sizeof(a));

    memset(ttr,,sizeof(ttr));

    memset(ltr,,sizeof(ltr));

    int m,n,num;

    char str[];

    scanf("%d%d",&m,&n);

    while(n--)

    {

        scanf("%d%s",&num,str);

        insert_tree(str,num);

    }

    scanf("%s",ttr);

    getBit();

    for(int i=; i<top; i+=)

    {

        int cc=;

        if(i+>top)continue;

        for(int j=i; j<i+; j++)

        {

            if(ltr[j]=='')

                cc++;

        }

        if((cc%==&&ltr[i+]=='')||(cc%==&&ltr[i+]==''))

        {

            for(int j=i; j<i+; j++)

            {

                a[tot++] = ltr[j]-'';

            }

        }

    }

    int root = ;

    for(int i=;i<tot;i++)

    {

        root = tree[root].next[a[i]];

        if(tree[root].num)

        {

            printf("%c",tree[root].num);

            m--;

            root = ;

        }

        if(!m)break;

    }

    puts("");

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        solve();

    }

    return ;

}

K.Supreme Number

A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers.

Now lets define a number NN as the supreme number if and only if each number made up of an non-empty subsequence of all the numeric digits of NN must be either a prime number or 1.

For example, 17 is a supreme number because 1, 7, 17 are all prime numbers or 1, and 19 is not, because 9 is not a prime number.

Now you are given an integer N (2≤N≤10^100), could you find the maximal supreme number that does not exceed NN?

Input

In the first line, there is an integer T≤100000 indicating the numbers of test cases.

In the following TT lines, there is an integer N (2≤N≤10^100).

Output

For each test case print "Case #x: y", in which xx is the order number of the test case and yy is the answer.

样例输入

样例输出

Case #1: 5

Case #2: 73

题目来源

ACM-ICPC 2018 沈阳赛区网络预赛

题意：找一个小于n的超级质数（其子序列也是质数），搜索。

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

int a[],top=;

int b[]= {,,,,},c[],d[];

int vis[];

int l,f;

void init()

{

    for(int i=; i<; i++)

        if(!vis[i])

            for(int j=i+i; j<; j+=i)

                vis[j]=;

}

void dfs1(int len,int x)

{

    if(len==l)

    {

        if(vis[x])

            f=;

        return ;

    }

    dfs1(len+,x*+c[len]);

    dfs1(len+,x);

}

int check(int x)

{

    f=;

    l=;

    while(x)

    {

        d[l++]=x%;

        x=x/;

    }

    for(int i=; i<l; i++)

        c[i]=d[l-i-];

    dfs1(,);

    return f;

}

void dfs(int len,int x)

{

    if(check(x)&&x)

        a[++top]=x;

    if(len==)

    {

        return ;

    }

    for(int i=; i<; i++)

        dfs(len+,x*+b[i]);

}

int main()

{

    int i,j,t,cas=,len,mmax,ans;

    char s[];

    init();

    dfs(,);

    sort(a+,a+top+);

    scanf("%d",&t);

    while(t--)

    {

        ++cas;

        scanf("%s",&s);

        len=strlen(s);

        if(len>)

            mmax=;

        else

        {

            mmax=;

            for(i=; i<len; i++)

                mmax=mmax*+(s[i]-'');

        }

        //printf("%d\n",mmax);

        for(i=; i<=top; i++)

            if(a[i]<=mmax)

                ans=a[i];

        printf("Case #%d: %d\n",cas,ans);

    }

    return ;

}