NOIP2013 题解

转圈游戏

　　题解：快速幂

 #include <cstdio>

 int n, m, k, x;

 inline long long QuickPow(int a, int k, int MOD){
     long long base = a, ret = ;
     while (k){
         if (k&) ret = (ret*base)%MOD;
         base = (base*base)%MOD, k >>= ;
     }
     return ret;
 }

 int main(){
     scanf("%d %d %d %d", &n, &m, &k, &x);
     printf("%lld\n", (x%n+(m%n)*QuickPow(, k, n)%n)%n);
 }

circle.cpp

火柴排队

　　题解：求逆序对(因为写不来归排就写的树状数组，怕TLE就蛋疼的加了一个读入优化)

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int MAXN =  +;
 const int MOD = ;

 int n, ans, c[MAXN], d[MAXN];
 char cc;

 struct Match{
     int v, n;

     friend bool operator < (const Match& A, const Match& B){
         return A.v<B.v;
     }
 }a[MAXN], b[MAXN];

 inline void add(int p, int v){
     while (p<=n)
         d[p] += v, p += (p&(-p));
 }

 inline int sum(int p){
     int ret = ;
     while (p>)
         ret += d[p], p -= (p&(-p));
     return ret;
 }

 inline int NextInt(){
     int ret = ;
     do cc = getchar();
     while (cc< || cc>);
     do ret *= , ret += (cc-), cc = getchar();
     while (<=cc && cc<=);
     return ret;
 }

 int main(){
     memset(d, , sizeof());
     n = NextInt(), ans = ;
     for (int i=; i<n; i++)
         a[i].v = NextInt(), a[i].n = i+;
     sort(a, a+n);
     for (int i=; i<n; i++)
         b[i].v = NextInt(), b[i].n = i+;
     sort(b, b+n);

     for (int i=; i<n; i++)
         c[a[i].n] = b[i].n;

     for (int i=; i<=n; i++)
         add(c[i], ),
         ans = (ans+i-sum(c[i]))%MOD;
     printf("%d\n", ans);
 }

match.cpp

货车运输

　　题解：先建一个最大生成树，再用lca乱搞一下就好了

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int MAXL = ;
 const int MAXN = +;
 const int MAXM = +;
 const int INF = 0x3f3f3f3f;

 int n, m, f[MAXN], nimo[MAXN][MAXL], an[MAXN][MAXL], dep[MAXN];
 bool vis[MAXN];
 char c;

 struct Edge{
     int d, to;
     Edge *next;
 }e[MAXM], *head[MAXN];

 struct Bary{
     int u, v, d;

     friend bool operator < (const Bary& A, const Bary& B){
         return A.d>B.d;
     }
 }bd[MAXM];

 inline int NextInt(){
     int ret = ;
     do c = getchar();
     while (c< || c>);
     do ret *= , ret += (c-), c = getchar();
     while (<=c && c<=);
     return ret;
 }

 inline int find(int x){
     return x==f[x] ? x : f[x]=find(f[x]);
 }

 int ne = ;
 inline void AddEdge(int f, int t, int d){
     e[ne].to = t, e[ne].d = d;
     e[ne].next = head[f];
     head[f] = e + ne++;
 }

 inline void Add(int u, int v, int d){
     f[find(u)] = find(v);
     AddEdge(u, v, d), AddEdge(v, u, d);
 }

 inline void BuildTree(int x){
     vis[x] = true;
     for (Edge *p=head[x]; p; p=p->next)
         if (!vis[p->to])
             an[p->to][] = x, nimo[p->to][] = p->d, dep[p->to] = dep[x]+, BuildTree(p->to);
 }

 inline int swim(int &x, int ht){
     int ret = INF;
     for (int i=MAXL-; i>=; i--)
         if (dep[an[x][i]]>=ht) ret = min(ret, nimo[x][i]), x = an[x][i];
     return ret;
 }

 inline int lca(int x, int y){
     int ret = INF;
     if (dep[x]^dep[y]) ret = dep[x]>dep[y] ? swim(x, dep[y]) : swim(y, dep[x]);
     if (x==y) return ret;
     for (int i=MAXL-; i>=; i--)
         if (an[x][i]^an[y][i])
             ret = min(ret, min(nimo[x][i], nimo[y][i])),
             x = an[x][i], y = an[y][i];
     return min(ret, min(nimo[x][], nimo[y][]));
 }

 int main(){
     memset(vis, false, sizeof(vis));
     memset(an, , sizeof(an));

     n = NextInt(), m = NextInt();
     for (int i=; i<=n; i++)
         f[i] = i;
     for (int i=; i<m; i++)
         bd[i].v = NextInt(), bd[i].u = NextInt(), bd[i].d = NextInt();

     sort(bd, bd+m);
     for (int i=; i<m; i++)
         if (find(bd[i].u)^find(bd[i].v)) Add(bd[i].u, bd[i].v, bd[i].d);
     for (int i=; i<=n; i++)
         if (!vis[i]) dep[i] = , BuildTree(i), nimo[i][] = INF, an[i][] = i;

     for (int i=; i<MAXL; i++)
         for (int j=; j<=n; j++)
             an[j][i] = an[an[j][i-]][i-],
             nimo[j][i] = min(nimo[j][i-], nimo[an[j][i-]][i-]);

     int Q, a, b;
     scanf("%d", &Q);
     while (Q--)
         scanf("%d %d", &a, &b),
         printf("%d\n", find(a)==find(b) ? lca(a, b) : -);
 }

truck.cpp

积木大赛：

　　题解：模拟，ans = ∑max{0, a[i]-a[i-1]}

 #include <cstdio>

 int num, h[];

 int max(int a,int b){
     return a>b ? a : b;
 }

 int main(){
     scanf("%d",&num);
     for (int i=; i<num; i++)
         scanf("%d", &h[i]);

     long long ans = ;
     for (int i=; i<num; i++)
         ans += max(, h[i]-h[i-]);
     printf("%lld",ans);
 }

block.cpp

花匠：

　　题解：贪心，缩点，把满足的连这的一群点缩为一个

 #include <cstdio>
 const int MAXN = 1e6+;

 int n, Pri, cj, a[MAXN];

 int main(){
     scanf("%d", &n), Pri = , cj = ;
     for (int i=; i<n; i++)
         scanf("%d", a+i);
     for (int i=; i<n; i++){
         if (a[i]>a[i-] && cj!=) cj = , Pri ++;
         if (a[i]<a[i-] && cj!=) cj = , Pri ++;
     }
     printf("%d", Pri);
 }

flower.cpp