Fighting For 2017 Season Contest 1

比赛地址【https://vjudge.net/contest/147011#problem/A】、960626

题目一：【http://codeforces.com/problemset/problem/701/A】、水题

题意：给出N张纸牌，N为偶数，每张纸牌上有数值，把这N张纸牌分给（N/2）个人，每个人分到两张纸牌，并且数值相同。一定存在解。

法一：

#include<bits/stdc++.h>
using namespace std;
int a[200], vis[200];
int N;
int main ()
{
    int sum = 0;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++)
    {
        scanf("%d", &a[i]);
        sum += a[i];
    }
    sum /= N / 2;
    for(int i = 1; i <= N; i++)
    {
        if(vis[i]) continue;
        vis[i] = 1;
        int t = sum - a[i];
        for(int j = i + 1; j <= N; j++)
        {
            if(vis[j]) continue;
            if(a[j] == t)
            {
                printf("%d %d
", i, j);
                vis[j] = 1;
                break;
            }
        }
    }
    return 0;
}

法二：

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int nu, id;
    bool operator <(const node x)const
    {
        return nu < x.nu;
    }
} A[150];
int N;
int main ()
{
    scanf("%d", &N);
    for(int i = 1; i <= N; i++)
    {
        scanf("%d", &A[i].nu);
        A[i].id = i;
    }
    sort(A + 1, A + N + 1);
    int s = 1, t = N;
    while(s <= t)
    {
        printf("%d %d
", A[s].id, A[t].id);
        s++, t--;
    }
    return 0;
}

题目二：【https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=614】

题意：

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a) if it is the empty string

(b) if A and B are correct, AB is correct,

(c) if A is correct, (A) and [A] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

题解：虽然长度最长只有128，但是是多组输入，用区间DP会TLE。简单stack的应用。

#include<stack>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF = 1e6;
const int MAXN = 150;
char S[MAXN];
int T, N;
int main ()
{
    scanf("%d", &T);
    getchar();
    while(T--)
    {
        gets(S + 1);
        N = strlen(S + 1);
        if(S[1] == ' ')//第一种情况
        {
            printf("Yes
");
            continue;
        }
        stack<char>st;
        for(int i = 1; i <= N; i++)
        {
            if(!st.empty() && ((st.top() == '(' && S[i] == ')') || (st.top() == '[' && S[i] == ']')))
                st.pop();//匹配
            else st.push(S[i]);
        }
        if(st.empty())    printf("Yes
");
        else    printf("No
");
    }
    return 0;
}

题目C:【http://codeforces.com/problemset/problem/382/C】/模拟

题意：给出一列数，使得加入一个数，使他是等差数列。从小到大输出可能的数，如果没有输出0，无限多个-1;

错了好多发。思维;

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 15;
map<int, int>mp;
int a[MAXN], N;
int main ()
{
    scanf("%d", &N);
    for(int i = 1; i <= N; i++)
        scanf("%d", &a[i]);
    sort(a + 1, a + N + 1);
    if(N == 1) printf("-1
");
    else if(N == 2)
    {
        if(a[1] == a[2]) printf("1
%d
", a[1]);
        else
        {
            int t = a[1] + a[2];
            if(t % 2)
                printf("2
%d %d
", 2 * a[1] - a[2], 2 * a[2] - a[1]);
            else
                printf("3
%d %d %d
", 2 * a[1] - a[2], t / 2, 2 * a[2] - a[1]);
        }
    }
    else
    {
        int t1 = -1, t2 = -1, p1, p2, nu = 0;
        for(int i = 1; i < N; i++)
        {
            int t = a[i + 1] - a[i];
            if(mp.count(t))  {mp[t]++;continue;}
            nu++;
            mp[t] = 1;
            if(nu == 1)
                t1 = t, p1 = i;
            else if(nu == 2)
                t2 = t, p2 = i;
            else break;
        }
        if(nu == 1)
        {
            if(t1 == 0)    printf("1
%d
", a[1]);
            else printf("2
%d %d
", a[1] - t1, a[N] + t1);
        }
        else if(nu == 2)
        {
            if(t1 > t2)
            {
                int t = a[p1] + a[p1 + 1];
                if(t1 != 2 * t2||mp[t1]!=1) printf("0
");
                else printf("1
%d
", t / 2);
            }
            else
            {
                int t = a[p2] + a[p2 + 1];
                if(t2 != 2 * t1||mp[t2]!=1) printf("0
");
                else printf("1
%d
", t / 2);
            }
        }
        else printf("0
");
    }
    return 0;
}

 题目四：【http://www.spoj.com/problems/INTSUB/en/】快速幂

题意：输入一个数n,有区间【1，2n】找出一个区间，区间中存在数a、b，使得a为区间中最小的数，b为a的整倍数数。

题解：枚举最小的数a,利用组合数公式求出所有的可能的情况，中间用快速幂处理。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 15;
const int mod = 1000000007;
int N;
LL fast_mod(int x)
{
    LL bas = 2, ans = 1;
    while(x)
    {
        if(x & 1)
            ans = ((ans % mod) * (bas % mod)) % mod;
        bas = ((bas % mod) * (bas % mod)) % mod;
        x >>= 1;
    }
    return ans % mod;
}
int main ()
{
    int T, ic = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &N);
        N <<= 1;
        LL ans = (fast_mod(N - 1) - 1 + mod) % mod;
        for(int i = 2; i <= N >> 1; i++)
        {
            LL t = N / i - 1;
            LL x = N - i - t;
            ans = (ans + (fast_mod(x) * ((fast_mod(t) - 1 + mod) % mod)) % mod) % mod;
        }
        printf("Case %d: %lld
", ++ic, ans);

    }
    return 0;
}

题目五：【http://codeforces.com/problemset/problem/607/B】

题意：给出n个数，每次删除一个回文串，求最少的删除次数。

题解：区间DP,dp[l][r],表示区间[l,r]中最少的回文串。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF=1e9;
const int MAXN = 550;
int dp[MAXN][MAXN];
int a[MAXN];
int N;
int main ()
{
    scanf("%d",&N);
    for(int i=1;i<=N;i++)
    {
        scanf("%d",&a[i]);
        dp[i][i]=1;
    }
    for(int l=2;l<=N;l++)
    {
        for(int i=1;i<=N-l+1;i++)
        {
            int j=i+l-1;
            dp[i][j]=INF;
            if(a[i]==a[j])
            {
                if(i+1==j)
                    dp[i][j]=dp[i+1][j-1]+1;
                else
                    dp[i][j]=dp[i+1][j-1];
            }
            for(int k=i;k<j;k++)
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
        }
    }
    printf("%d
",dp[1][N]);
    return 0;
}