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(); //} }