codeforces 497B Tennis Game 思维+二分~

链接：http://codeforces.com/problemset/problem/497/B

这种题考察的是基本功。。像我这种基本功为0的喳喳只能卡在这里了。。。

思路：n范围为10^5, 必须找出nlogn的算法才行。枚举t,然后用二分找出解。。巧妙~~ 好题赞一个，虽然不会做~~~

代码：

/* ***********************************************
Author        :ltwy
Created Time  :2014年12月18日 星期四 16时28分12秒
File Name     :1.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

int n;
int a[101000];
int sum1[101000];
int sum2[101000];

pair<int, int> p[100100];

int get_next(int s, int st, int ed)
{
    if(sum1[ed] - sum1[st-1] <s && sum2[ed] - sum2[st-1] < s)
        return -1;
    int l = st, r = ed;
    int id = ed;

    while(l <= r)
    {
        int mid = (l+r)>>1;
        if(sum1[mid] - sum1[st-1] < s&& sum2[mid] - sum2[st-1] < s)
        {
            l = mid+1;
        }
        else
        {
            id = mid;
            r = mid -1;
        }
    }
    return id;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    scanf("%d", &n);

    for(int i=1; i<=n; i++)
    {
        scanf("%d", &a[i]);
    }

    memset(sum1, 0, sizeof(sum1));
    memset(sum2, 0, sizeof(sum2));

    int T1 = 0, T2 = 0;
    for(int i = 1; i<=n; i++)
    {
        if(a[i] == 1)
            T1++;
        if(a[i] == 2)
            T2++;
        sum1[i] = T1;
        sum2[i] = T2;
    }

    int cnt = 0;
    int W1 = 0;
    int W2 = 0;
    bool flag;
    for(int s=1; s<=n; s++)
    {
        W1 = 0;
        W2 = 0;
        flag = false;
        if(sum1[n] < s && sum2[n] < s)
            break;
        int pos = 1;
        int last = 0;
        while(pos <= n)
        {
            int temp = get_next(s, pos, n);
            if(temp == -1)
            {
                flag = true;
                break;
            }
            if(a[temp] == 1)
            {
                W1++;
                last = 1;
            }
            else
            {
                W2++;
                last = 2;
            }
            pos = temp+1;
        }
        if(flag)
            continue;
        if(W1 == W2)
            continue;
        if(W1 > W2&& last == 2)
            continue;
        if(W2 > W1&& last == 1)
            continue;
        p[cnt++] = make_pair(max(W1, W2), s);
    }

    printf("%d\n", cnt);
    sort(p, p+cnt);
    for(int i=0; i<cnt; i++)
        printf("%d %d\n", p[i].first, p[i].second);
    return 0;
}