UESTC 2014 summer training #10 div.2

B. Race to 1 ultraviolet A 11762

The first contact with probability DP, I did not expect it to be DP...

First, you need to know the total expectation = the expectation of each thing × the probability of each thing happening

Then we can write a recursive formula based on this, which is also DP?

Assume that there are m prime numbers not less than X, and N qualitative factors of X (more accurate), then when X is expected, we can consider that X, X/pi p is the prime factor of X.

Therefore, it is not difficult to conclude that E (x) = 1 + (m-N)/ME (x) + 1/msigmae (x/PI) will add 1, because X is transferred to the back end, there is one more step.

// Updated: every case is run out, and the F array does not need memset... it will be 10 times faster! In the case

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;const int maxn = 1000000+50;const double eps = 1e-10;int T, N, tot;int prime[maxn];double f[maxn];bool flag[maxn], done[maxn];void getPrime(){    for(int i = 2; i < maxn; i++) {        if(!flag[i])    prime[tot++] = i;        for(int j = 0; j < tot && prime[j]*i < maxn; j++) {            flag[i*prime[j]] = true;            if(i % prime[j] == 0)    break;        }    }}double dp(int x){    if(done[x])    return f[x];    done[x] = true;    int n, m;    n = m = 0;    for(int i = 0; i < tot && prime[i] <= x; i++) {        n++;        if(x % prime[i] == 0) {            m++;            f[x] += dp(x/prime[i]);        }    }//    if(x == N)    cout << n << ‘ ‘ << m << endl;    return f[x] = (f[x]+n)/m;}int main(){#ifdef LOCAL    freopen("B.in", "r", stdin);#endif    getPrime();    scanf("%d", &T);    for(int t = 1; t <= T; t++) {        scanf("%d", &N);        memset(done, 0, sizeof(done));        memset(f, 0, sizeof(f));        f[1] = 0.0;    done[1] = true;        dp(N);        printf("Case %d: %.10lf\n", t, f[N]);    }    return 0;}

 

D. Jumping Mario ultraviolet (a) 11764

#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;int T, n, now, high, low;int main(){    cin >> T;    for(int t = 1; t <= T; t++) {        high = 0;        low = 0;        cin >> n;        for(int i = 0; i < n; i++) {            int x;            cin >> x;            if(i == 0) {                now = x;                continue;            }            if(x > now)    high++;            else if(x < now)    low++;            now = x;        }        cout << "Case " << t << ": " << high << ‘ ‘ << low << endl;    }    return 0;}

 

J. Lighting away ultraviolet A 11770

The first reaction is indeed a strongly connected component... contraction point... but I am not familiar with... so I went to the yydfs method, but I was not enough IQ... I didn't debug it.

The positive solution is to reduce the point dyeing to determine the new graph entry into the 0 point.

Debugging for a long time... all kinds of small errors have been found, and finally fell out of the stack and forgot to increase the brackets...

 

#include <iostream>#include <cstdio>#include <cstdlib>#include <stack>#include <cstring>using namespace std;const int maxn = 100000+50;int T, n, m, idx, tot, cnt;int head[maxn], next[maxn], edge[maxn], dfn[maxn], low[maxn], in[maxn], type[maxn];bool instack[maxn], visit[maxn];stack<int> st;void add(int u, int v){    edge[tot] = v;    next[tot] = head[u];    head[u] = tot++;}void tarjan(int u){    dfn[u] = low[u] = ++idx;    st.push(u);    instack[u] = true;    visit[u] = true;    for(int e = head[u]; e != -1; e = next[e]) {        int v = edge[e];        if(!visit[v]) {            tarjan(v);            low[u] = min(low[u], low[v]);        }                else if(instack[v])            low[u] = min(low[u], dfn[v]);    }    if(dfn[u] == low[u]) {        cnt++;        while(!st.empty())        {            int now=st.top();    st.pop();            instack[now]=0;            type[now]=cnt;            if(now == u) break;        }    }}int main(){#ifdef LOCAL    freopen("J.in", "r", stdin);#endif    scanf("%d", &T);    for(int t = 1; t <= T; t++) {        scanf("%d%d", &n, &m);        memset(next, -1, sizeof(next));        memset(head, -1, sizeof(head));        memset(in, 0, sizeof(in));        memset(instack, 0, sizeof(instack));        memset(visit, 0, sizeof(visit));        tot = 0;    idx = 0;    cnt = 0;        for(int i = 0; i < m; i++) {            int u, v;            scanf("%d%d", &u, &v);            add(u, v);        }//        for(int i = 1; i <= n; i++)//            if(!visit[i] && in[i] == 0)    tarjan(i);        for(int i = 1; i <= n; i++)             if(!visit[i])    tarjan(i);        int ans = 0;        for(int i = 1; i <= n; i++)            for(int e = head[i]; e != -1; e = next[e]) {                if (type[i] != type[edge[e]]) in[type[edge[e]]]++;            }        for (int i = 1;i <= cnt;i++)            if (in[i] == 0) ans++;//        for(int i = 1; i <= n; i++)//            if(in[i] == 0)    cnt++;        printf("Case %d: %d\n", t, ans);    }    return 0;}

 

During the competition, I only got a question... then J and B are really not strong enough... J is a bit confused. Now I understand it.

 

UESTC 2014 summer training #10 div.2