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 = 1;
    while (k){
        if (k&1) ret = (ret*base)%MOD;
        base = (base*base)%MOD, k >>= 1;
    }
    return ret;
}

int main(){
    scanf("%d %d %d %d", &n, &m, &k, &x);
    printf("%lld
", (x%n+(m%n)*QuickPow(10, k, n)%n)%n);
}

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 =  100000+10;
const int MOD = 99999997;

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 = 0;
    while (p>0)
        ret += d[p], p -= (p&(-p));
    return ret;
}

inline int NextInt(){
    int ret = 0;
    do cc = getchar();
    while (cc<48 || cc>57);
    do ret *= 10, ret += (cc-48), cc = getchar();
    while (48<=cc && cc<=57);
    return ret;
}

int main(){
    memset(d, 0, sizeof(0));
    n = NextInt(), ans = 0;
    for (int i=0; i<n; i++)
        a[i].v = NextInt(), a[i].n = i+1;
    sort(a, a+n);
    for (int i=0; i<n; i++)
        b[i].v = NextInt(), b[i].n = i+1;
    sort(b, b+n);
    
    for (int i=0; i<n; i++)
        c[a[i].n] = b[i].n;
    
    for (int i=1; i<=n; i++)
        add(c[i], 1), 
        ans = (ans+i-sum(c[i]))%MOD;
    printf("%d
", ans);
}

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 = 20;
const int MAXN = 10000+10;
const int MAXM = 50000+10;
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 = 0;
    do c = getchar();
    while (c<48 || c>57);
    do ret *= 10, ret += (c-48), c = getchar();
    while (48<=c && c<=57);
    return ret;
}

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

int ne = 0;
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][0] = x, nimo[p->to][0] = p->d, dep[p->to] = dep[x]+1, BuildTree(p->to);
}

inline int swim(int &x, int ht){
    int ret = INF;
    for (int i=MAXL-1; i>=0; 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-1; i>=0; 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][0], nimo[y][0]));
}

int main(){
    memset(vis, false, sizeof(vis));
    memset(an, 0, sizeof(an));

    n = NextInt(), m = NextInt();
    for (int i=1; i<=n; i++)
        f[i] = i;
    for (int i=0; i<m; i++)
        bd[i].v = NextInt(), bd[i].u = NextInt(), bd[i].d = NextInt();
    
    sort(bd, bd+m);
    for (int i=0; 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=1; i<=n; i++)
        if (!vis[i]) dep[i] = 0, BuildTree(i), nimo[i][0] = INF, an[i][0] = i;

    for (int i=1; i<MAXL; i++)
        for (int j=1; j<=n; j++)
            an[j][i] = an[an[j][i-1]][i-1],
            nimo[j][i] = min(nimo[j][i-1], nimo[an[j][i-1]][i-1]);
    
    int Q, a, b;
    scanf("%d", &Q);
    while (Q--)
        scanf("%d %d", &a, &b),
        printf("%d
", find(a)==find(b) ? lca(a, b) : -1);
}

积木大赛：

　　题解：模拟，ans = ∑max{0, a[i]-a[i-1]}

#include <cstdio>

int num, h[100010];

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

int main(){
    scanf("%d",&num);
    for (int i=0; i<num; i++)
        scanf("%d", &h[i]);
    
    long long ans = 0;
    for (int i=0; i<num; i++)
        ans += max(0, h[i]-h[i-1]);
    printf("%lld",ans);
}

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+10;

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

int main(){
    scanf("%d", &n), Pri = 1, cj = 1221;
    for (int i=0; i<n; i++)
        scanf("%d", a+i);
    for (int i=1; i<n; i++){
        if (a[i]>a[i-1] && cj!=1) cj = 1, Pri ++;
        if (a[i]<a[i-1] && cj!=0) cj = 0, Pri ++;
    }
    printf("%d", Pri);
}