Mango Weekly Training Round #3 解题报告

A. Codeforces 92A Chips

签到题。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 10007

int a[55];

int main()
{
    int n,m,i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
            a[i] = i;
        i = 1;
        while(1)
        {
            if(m>=a[i])
                m-=a[i];
            else
                break;
            i++;
            if(i == n+1)
                i = 1;
        }
        cout<<m<<endl;
    }
    return 0;
}

B.Codeforces 217A Ice Skating

dfs或者并查集。

dfs：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 107

struct node
{
    int x,y;
}a[N];
int vis[N];
int n;

void joint(int k)
{
    if(!vis[k])
    {
        vis[k] = 1;
        for(int i=0;i<n;i++)
        {
            if(i != k && (a[k].x == a[i].x || a[k].y == a[i].y))
                joint(i);
        }
    }
}

int main()
{
    int i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        memset(vis,0,sizeof(vis));
        int cnt = 0;
        for(i=0;i<n;i++)
        {
            if(!vis[i])
            {
                cnt++;
                joint(i);
            }
        }
        cout<<cnt-1<<endl;
    }
    return 0;
}

C.UVA 12592 Slogan Learning of Princess

hash,用map做就可以了，也可以用字符串数组

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
#define N 10007

map<string,string> mp;

int main()
{
    int n,i,q;
    string s1,s2;
    while(scanf("%d",&n)!=EOF)
    {
        getchar();
        for(i=0;i<n;i++)
        {
            getline(cin,s1);
            cin.clear();
            getline(cin,s2);
            cin.clear();
            mp[s1] = s2;
        }
        scanf("%d",&q);
        getchar();
        while(q--)
        {
            getline(cin,s1);
            cout<<mp[s1]<<endl;
        }
    }
    return 0;
}

D.HDU 4027 Can you answer these queries?

线段树，单点更新，用一个标记表示区间内是否全为1，全为1则不用更新，以节省操作时间。注意：要用_int64类型，不要忘记每组数据后打一个空行

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define lll __int64
using namespace std;
#define N 100007

struct node
{
    lll sum;
    int mark;
}tree[4*N];

void pushup(int rt)
{
    tree[rt].sum = tree[2*rt].sum + tree[2*rt+1].sum;
    tree[rt].mark = tree[2*rt].mark && tree[2*rt+1].mark;
}

void build(int l,int r,int rt)
{
    tree[rt].mark = 0;
    if(l == r)
    {
        scanf("%I64d",&tree[rt].sum);
        if(tree[rt].sum == 1)
            tree[rt].mark = 1;
        return;
    }
    int mid = (l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
    pushup(rt);
}

void update(int l,int r,int aa,int bb,int rt)
{
    if(aa<=l && bb>=r)
    {
        if(tree[rt].mark)
            return;
        if(!tree[rt].mark && l == r)
        {
            tree[rt].sum = (lll)sqrt(tree[rt].sum);
            if(tree[rt].sum <= 1)
                tree[rt].mark = 1;
            return;
        }
    }
    int mid = (l+r)/2;
    if(aa<=mid)
        update(l,mid,aa,bb,2*rt);
    if(bb>mid)
        update(mid+1,r,aa,bb,2*rt+1);
    pushup(rt);
}

lll query(int l,int r,int aa,int bb,int rt)
{
    if(aa>r || bb<l)
        return 0;
    if(aa<=l && bb>=r)
        return tree[rt].sum;
    int mid = (l+r)/2;
    return query(l,mid,aa,bb,2*rt)+query(mid+1,r,aa,bb,2*rt+1);
}

int main()
{
    int n,i,q,op,aa,bb,cs = 1;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,n,1);
        scanf("%d",&q);
        printf("Case #%d:
",cs++);
        while(q--)
        {
            scanf("%d%d%d",&op,&aa,&bb);
            if(aa>bb)
                swap(aa,bb);
            if(op)
                printf("%I64d
",query(1,n,aa,bb,1));
            else
                update(1,n,aa,bb,1);
        }
        printf("
");
    }
    return 0;
}

E.UVA 11488 Hyper Prefix Sets

字典树，结构node维护两个值： count 和 deep ，结果即为节点的count * deep 的最大值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 50007

struct node
{
    int count,deep;
    node *next[2];
}*root;

char ss[N];
int maxi;

node *create()
{
    node *p;
    p = (node *)malloc(sizeof(node));
    p->count = 0;
    p->deep = 0;
    for(int i=0;i<2;i++)
        p->next[i] = NULL;
    return p;
}

void release(node *p)
{
    for(int i=0;i<2;i++)
    {
        if(p->next[i] != NULL)
            release(p->next[i]);
    }
    free(p);
}

void insert(char *ss)
{
    node *p = root;
    int i = 0,k;
    while(ss[i])
    {
        k = ss[i++] - '0';
        if(p->next[k] == NULL)
            p->next[k] = create();
        p->next[k]->deep = p->deep + 1;
        p = p->next[k];
        p->count++;
        maxi = max(maxi,p->count*p->deep);
    }
}

int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--)
    {
        root = create();
        scanf("%d",&n);
        maxi = -1000000;
        for(i=0;i<n;i++)
        {
            scanf("%s",ss);
            insert(ss);
        }
        cout<<maxi<<endl;
        release(root);
    }
    return 0;
}

F.UVALive 6655 Two Points Revisited

构造法。