2014编程之美初赛第一场题解

第一题：http://hihocoder.com/contest/msbop2014r2a/problem/1

直接打公式即可，可以用strtod和string减少工作量，选错gcc了，贡献了一个ce...........

// #pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <iterator>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
#define ROUND(x) round(x)
#define FLOOR(x) floor(x)
#define CEIL(x) ceil(x)
const int maxn = 10010;
const int maxm = 0;
const int inf = 0x3f3f3f3f;
const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double INF = 1e30;
const double eps = 1e-6;
const int P[4] = {0, 0, -1, 1};
const int Q[4] = {1, -1, 0, 0};
const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
int kase;
char str1[maxn], str2[maxn], str3[maxn];
void init()
{
    kase++;
}
void input()
{
    scanf("%s%s%s", str1, str2, str3);
}
void debug()
{
    //
}
double f1, f2, f3;
void solve()
{
    string tmp1, tmp2;
    int strn = strlen(str1);
    tmp1.clear();
    tmp2.clear();
    for (int i = 0; i < strn; i++)
    {
        if (str1[i] == '.' || (str1[i] >= '0' && str1[i] <= '9'))
        {
            tmp1.push_back(str1[i]);
        }
        else tmp2.push_back(str1[i]);
    }
    f1 = strtod(tmp1.c_str(), NULL);
    if (tmp2 == "m") f1 *= 1000.0;
    else if (tmp2 == "dm") f1 *= 100.0;
    else if (tmp2 == "cm") f1 *= 10.0;
    else if (tmp2 == "dm") f1 *= 100.0;
    else if (tmp2 == "um") f1 /= 1000.0;
    else if (tmp2 == "nm") f1 /= 1000.0 * 1000.0;

    strn = strlen(str2);
    tmp1.clear();
    tmp2.clear();
    for (int i = 0; i < strn; i++)
    {
        if (str2[i] == '.' || (str2[i] >= '0' && str2[i] <= '9'))
        {
            tmp1.push_back(str2[i]);
        }
        else tmp2.push_back(str2[i]);
    }
    f2 = strtod(tmp1.c_str(), NULL);
    if (tmp2 == "m") f2 *= 1000.0;
    else if (tmp2 == "dm") f2 *= 100.0;
    else if (tmp2 == "cm") f2 *= 10.0;
    else if (tmp2 == "dm") f2 *= 100.0;
    else if (tmp2 == "um") f2 /= 1000.0;
    else if (tmp2 == "nm") f2 /= 1000.0 * 1000.0;

    strn = strlen(str3);
    tmp1.clear();
    tmp2.clear();
    for (int i = 0; i < strn; i++)
    {
        if (str3[i] == '.' || (str3[i] >= '0' && str3[i] <= '9'))
        {
            tmp1.push_back(str3[i]);
        }
        else tmp2.push_back(str3[i]);
    }
    f3 = strtod(tmp1.c_str(), NULL);

    double ans = f3 * (f1 / f2);

    printf("Case %d: %.2fpx
", kase, ans);
}
void output()
{
    //
}
int main()
{
    // std::ios_base::sync_with_stdio(false);
    // #ifndef ONLINE_JUDGE
    // freopen("in.cpp", "r", stdin);
    // #endif
    int T;
    scanf("%d", &T);
    kase = 0;
    while (T--)
    {
        init();
        input();
        solve();
        output();
    }
    return 0;
}

第二题：http://hihocoder.com/contest/msbop2014r2a/problem/2

神奇地暴力过了。。。。。。。。听说正解是用树状数组

dfs，第一次求深度，然后每次操作从u点dfs一遍(记录其fa节点即可直接从u点dfs)，可以加入剪枝(dep>r的剪掉)，注意边加delta边模防溢出

注意算最终的结果的时候要边乘边模，否则会溢出，而且乘magic的时候用long long

// #pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <iterator>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
#define ROUND(x) round(x)
#define FLOOR(x) floor(x)
#define CEIL(x) ceil(x)
const int maxn = 1e5 + 10;
const int maxm = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double INF = 1e30;
const double eps = 1e-6;
const int MOD = 1000000007;
const int MAGIC = 12347;
const int P[4] = {0, 0, -1, 1};
const int Q[4] = {1, -1, 0, 0};
const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
struct Node
{
    int dep;
    int val;
    Node(int _dep = 0, int _val = 0): dep(_dep), val(_val) {}
} node[maxn];
struct Edge
{
    int u, v;
    int next;
} edge[maxm];
int head[maxn], en;
int fa[maxn];
int n;
void addSubEdge(int u, int v)
{
    edge[en].u = u;
    edge[en].v = v;
    edge[en].next = head[u];
    head[u] = en++;
}
void addEdge(int u, int v)
{
    addSubEdge(u, v);
    addSubEdge(v, u);
}
int kase;
void init()
{
    memset(head, -1, sizeof(head));
    en = 0;
    kase++;
}
void dfs1(int u, int p)
{
    if (u != 0) node[u].dep = node[p].dep + 1;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if (v == p) continue;
        dfs1(v, u);
    }
}
void input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++) node[i] = Node();
    node[0].dep = 1;
    fa[0] = 0;
    for (int i = 1; i < n; i++)
    {
        int tt;
        scanf("%d", &tt);
        tt--;
        addEdge(tt, i);
        fa[i] = tt;
    }
    dfs1(0, -1);
}
void debug()
{
    //
}
void dfs(int u, int p, int l, int r, int w)
{
    if (node[u].dep <= r && node[u].dep >= l)
    {
        node[u].val += w;
        node[u].val %= MOD;
    }
    if (node[u].dep == r) return;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if (v == p) continue;
        if (node[v].dep > r) continue;
        dfs(v, u, l, r, w);
    }
}
void solve()
{
    int Q;
    scanf("%d", &Q);
    int u, l, r, w;
    while (Q--)
    {
        scanf("%d%d%d%d", &u, &l, &r, &w);
        u--;
        dfs(u, fa[u], l, r, w);
    }
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
        // cout << node[i].val << endl;
        LL tmp = (LL)ans * (LL)MAGIC;
        // cout << "tmp:" << tmp << endl;
        tmp %= (LL)MOD;
        ans = (int)tmp;
        ans %= MOD;
        // cout << ans << endl;
        ans += node[i].val;
        ans %= MOD;
        // cout << ans << endl;
    }
    printf("Case %d: %d
", kase, ans);
}
void output()
{
    //
}
int main()
{
    // std::ios_base::sync_with_stdio(false);
    // #ifndef ONLINE_JUDGE
    // freopen("in.cpp", "r", stdin);
    // #endif

    int T;
    scanf("%d", &T);
    kase = 0;
    while (T--)
    {
        init();
        input();
        solve();
        output();
    }
    return 0;
}

第三题：http://hihocoder.com/contest/msbop2014r2a/problem/3

三分法，显然要求的点x到其他所有点的距离之和随x轴先减后增，符合三分法

// #pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <iterator>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
#define ROUND(x) round(x)
#define FLOOR(x) floor(x)
#define CEIL(x) ceil(x)
const int maxn = 1e5 + 10;
const int maxm = 0;
const int inf = 0x3f3f3f3f;
const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double INF = 1e30;
const double eps = 1e-10;
const int P[4] = {0, 0, -1, 1};
const int Q[4] = {1, -1, 0, 0};
const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
int kase;
int xx[maxn], yy[maxn];
int n;
void init()
{
    kase++;
}
void input()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d%d", &xx[i], &yy[i]);
}
void debug()
{
    //
}
double dis(double ax, double ay, double bx, double by)
{
    return sqrt((ax - bx) * (ax - bx) + (by - ay) * (by - ay));
}
double cal(double nx)
{
    double ans = 0;
    for (int i = 0; i < n; i++)
        ans += dis(nx, 0, xx[i], yy[i]);
    return ans;
}
void solve()
{
    double Left, Right;
    double mid, midmid;
    double mid_area, midmid_area;
    Left = 0;
    Right = 1e6;
    while (Left + eps < Right)
    {
        mid = (Left + Right) / 2;
        midmid = (mid + Right) / 2;
        mid_area = cal(mid);
        midmid_area = cal(midmid);
        if (mid_area < midmid_area) Right = midmid;
        else Left = mid;
    }
    printf("Case %d: %.8f
", kase, Left);
}
void output()
{
    //
}
int main()
{
    // std::ios_base::sync_with_stdio(false);
    // #ifndef ONLINE_JUDGE
    // freopen("in.cpp", "r", stdin);
    // #endif
    int T;
    scanf("%d", &T);
    kase = 0;
    while (T--)
    {
        init();
        input();
        solve();
        output();
    }
    return 0;
}