最少说服人数（二分+贪心）

http://codeforces.com/gym/101755/problem/K

题意：n个人，需要m个人评论，从头开始依次采访，每一个人需要ai个人评论才评论，或说服他去评论，问最少需要说服几个人？

解法：二分说服评论人数，因为说服评论人数对于条件来说是单调性，说服越多越能满足条件。

check贪心策略：如果当前评论人数达到要求，不需说服该人就会评论，如果没达到要求则能说服则说服。因为说服越早对后面越有利。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 500009 ;
int a[200009];
int n , p ;

int check(int k)//代表可说服人数
{
    int ans = 0 ;//
    for(int i = 0 ; i < n ; i++)
    {
        if(ans >= a[i])
        {
            ans++;
        }
        else if(k)//不符合，能说服就说服
        {
            k--;
            ans++ ;
        }
        if(ans == p)
            return 1 ;
    }
    return 0 ;
}


int main()
{

    while(~scanf("%d%d" , &n ,&p))
    {
        for(int i = 0 ; i < n ; i ++)
        {
            scanf("%d"
                  
                   , &a[i]);
        }
        int l = 0 , r = p , mid , ans = p ;
        while(l <= r)
        {
            mid = (l + r) / 2 ;
            if(check(mid))//如果答案是有mid个人评论了，判断是否符合题意
            {
                r = mid - 1 ;//如果mid个人评论符合题意，区间右区间就左移，继续找最小答案
                ans = mid ;
            }
            else
                l = mid + 1 ;
        }
        cout << ans << endl ;
    }


    return 0 ;
}

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define int ll
using namespace std;
const int maxn = 2e5+9;
int a[maxn];
int n , m ;

bool check(int x){
    int p = x ;
    int view = 0;
    rep(i , 1 , n){
        //cout << x << " " << i << " " << view << " " << a[i] << endl;
        if(view >= a[i]){
            view++;
        }else if(p > 0){
            view++;
            p--;
        }
    }
    return view >= m ;
}

void solve(){
    cin >> n >> m ;
    rep(i , 1 , n){
        cin >> a[i] ;
    }
    int l = 0 , r = m ;
    while(l <= r-3){
        int mid = (l+r) >> 1;
        if(check(mid)){
            r = mid ;
        }else{
            l = mid ;
        }
    }
    rep(i , l , r){
        if(check(i)){
            cout << i << endl;
            break ;
        }
    }

}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    //int t ;
    //cin >> t ;
    //while(t--){
    //while(~scanf("%s%lld" , s+1 , &n))
        solve();
    //}
}