http://codeforces.com/problemset/problem/348/A

題意：有n個(gè)人，在玩一個(gè)游戲，游戲表示每局都必須有個(gè)管理員參加

告訴你，每個(gè)人想當(dāng)多少局玩家，然后讓你求，最少多少局游戲，才能滿足題意！

C++版本一

題解：貪心+公式

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; ll a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%I64d",&a[i]),sum+=a[i],ans=max(a[i],ans);n--;temp=sum/n;if(sum%n)temp++;cout<<max(ans,temp)<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

C++版本二

題解：二分

如果可以進(jìn)行游戲，那么必定每次游戲都會(huì)有一個(gè)人當(dāng)裁判。所以判斷是否可以進(jìn)行游戲的條件就是檢查對(duì)于n次游戲，是否都有n個(gè)人充當(dāng)裁判。

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const ll INF = 0x3f3f3f3f3f3f3f; int t,n,m,k,p; ll l,r,u,v; ll ans,cnt,flag,temp,sum; ll a[N]; bool sloved(ll num){ll cnt=0;for(int i=1;i<=n;i++){if(num<a[i])return false;cnt+=num-a[i];}return num<=cnt; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);l=0;r=1e12;while(l<=r){ll mid=(l+r)>>1;if(sloved(mid)){ans=mid;r=mid-1;}else{l=mid+1;}}cout<<ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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