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A brief description of the simulation game

The title can be found on the Loj.

10178."5.5 cases of 4" travel problem

Simple problem, expand N to 2 * N, monotone queue can be, pay attention to the positive and negative.

#include <iostream> #include <cstring> #include <cmath> #include <cstdio> #include <  Algorithm> using namespace std;  typedef long Long LL;    const int N=2000005;LL sum[n];int p[n],d[n],que[n],ok[n];int n;inline int read () {int x = 0,f = 1;char c = GetChar ();    while (C < ' 0 ' | | c > ' 9 ') {if (c = = '-') F = -1;c = GetChar ();}    while (c >= ' 0 ' && C <= ' 9 ') {x = x * + C-' 0 '; c = GetChar ();} return x * f;}    void DP1 () {int q1 = 1,Q2 = 0;        for (int i = 1; I <= n; + + i) {while (Q1 <= Q2 && SUM[QUE[Q2]] >= sum[i])--Q2;    que[++ q2]=i;        } for (int i = n + 1; I <= n * 2, + + i) {while (Q1 <= Q2 && SUM[QUE[Q2]] >= sum[i])--Q2;        que[++ Q2] = i;        while (Q1 <= Q2 && QUE[Q1] <= i-n) + + Q1;        if (sum[que[q1]] >= sum[i-n]) ok[i-n] = 1;    }}void DP2 () {int q1 = 1,Q2 = 0; for (int i = n * 2, i > N;---I) {while (Q1 <= Q2 && SUM[QUE[Q2]] >= sum[i])--Q2;    que[++ Q2] = i;        } for (int i = n; I,--i) {while (Q1 <= Q2 && SUM[QUE[Q2]] >= sum[i])--Q2;        que[++ Q2] = i;        while (Q1 <= Q2 && QUE[Q1] >= i + N) + + Q1;          if (sum[que[q1]] >= sum[i + n]) ok[i] = 1;    }}int Main () {n = read ();         for (int i = 1;i <= n;++ i) {p[i + n] = p[i] = Read (), D[i + n] = d[i] = read ();     Sum[i] = Sum[i-1] + p[i-1]-d[i-1];    } for (int i = n + 1;i <= 2 * n;++ i) sum[i] = Sum[i-1] + p[i-1]-d[i-1];    DP1 ();    for (int i = n * 2; i;--i) sum[i] = sum[i + 1] + p[i + 1]-d[i];    DP2 (); for (int i = 1;i <= n;++ i) ok[i]?    Puts ("TAK"): Puts ("NIE");   return 0;}

10078."3.2 Practice 4" good New Year

Simple problem, SPFA run six-side shortest way, full array enumeration can be.

#include <iostream> #include <cstring> #include <queue> #include <cstdio>const int maxn = 5e4 + 7; const int MAXM = 1e5 + 7;using namespace std;int ke_dis[maxn][7];queue <int>q;bool vis[maxn];int dis[maxn];int Q[7], N,m; struct Node {int v,nex,w;}    MAP[MAXM << 1];int head[maxn],num;void add_node (int u,int v,int w) {map[++ num] = (Node) {v,head[u],w};    HEAD[U] = num; return;}    inline int read () {int x = 0,f = 1;char c = GetChar ();    while (C < ' 0 ' | | c > ' 9 ') {if (c = = '-') F = -1;c = GetChar ();}    while (c >= ' 0 ' && C <= ' 9 ') {x = x * + C-' 0 '; c = GetChar ();} return x * f;}    void spfa (int beg,int bh) {memset (dis,0x3f,sizeof (dis));    Dis[beg] = 0;    Q.push (Beg);    Vis[beg] = true;        while (!q.empty ()) {int p = Q.front (); Q.pop (); vis[p] = false;            for (int i = Head[p];i;i = Map[i].nex) {int v = MAP[I].V; if (Dis[v] > Dis[p] + map[i].w) {Dis[v] = Dis[p] + MAP[I].W ;                    if (!vis[v]) {Q.push (v);                VIS[V] = true;    }}}} for (int i = 1;i <= n;++ i) ke_dis[i][bh] = dis[i]; return;} Long Long Ans;bool Is_ch[7];int ch[7];//Five relatives walk in order, just numbered q[number] is station number//special q[0] = 1 relative should be 1 void work () {Long Long sum    = 0;    for (int i = 1;i <= 5;++ i) {sum + = (long Long) ke_dis[q[ch[i-1]]][ch[i]]; } ans = min (ans,sum);    } void Dfs (int tot) {if (tot = = 6) work ();            for (int i = 1;i <= 5;++ i) {if (!is_ch[i]) {Ch[tot] = i;is_ch[i] = true;            DFS (tot + 1);        Is_ch[i] = false;    }}}int Main () {n = read (); m = read ();    Q[0] = 1;    for (int i = 1;i <= 5;++ i) q[i] = read ();        for (int i = 1,u,v,w;i <= m;++ i) {u = read (); v = Read (); w = read ();            Add_node (U,V,W);    Add_node (V,U,W);    } for (int i = 0;i <= 5;++ i) SPFA (q[i],i);    ans = 1e12; DFS (1);   printf ("%lld", ans); return 0;}

#10220. "One-pass 6.5 cases 2"fibonacci nth

The simplest of topics.
Set a matrix multiplication can be.

#include <iostream>#include <cstring>#include <cstdio>#define ll long longusing namespace std;struct Node {    ll w[4][4];    Node() {        memset(w,0,sizeof(w));    }};ll n,m;Node mul(Node a,Node b) {    Node c;    for(ll i = 1;i <= 2;++ i) {        for(ll j = 1;j <= 2;++ j) {            for(ll k = 1;k <= 2;++ k) {                c.w[i][j] = ( c.w[i][j] + a.w[i][k] * b.w[k][j] ) % m;            }        }    }    return c;}void fast_pow(ll b) {    Node a,ans;    a.w[1][1] = a.w[1][2] = a.w[2][1] = 1;    ans.w[1][1] = ans.w[2][2] = 1;    for(;b;b >>= 1,a = mul(a,a) ) {        if(b & 1) ans = mul(ans,a);    }    Node tmp;    tmp.w[1][1] = tmp.w[1][2] = 1;    ans = mul(tmp,ans);    printf("%lld", ans.w[1][1]);    return ;}int main() {    scanf("%lld%lld",&n,&m);    if(n == 1) {puts("1");return 0;}    if(n == 2) {puts("1");return 0;}    fast_pow(n - 2);    return 0;}

Test Harvest

The topic must be read well, read the accurate.
Otherwise, like this time, \ (220\)
Violence must be fought.
otherwise \ (250\) points of violence.
Improve your own code and error-checking levels
Now, Gzy I'm going to start serious now.

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