Codeforces Round #202 (Div. 1) A. Mafia
A. Mafiatime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output
One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n?-?1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play ai rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
InputThe first line contains integer n (3?≤?n?≤?105). The second line contains n space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?109) — the i-th number in the list is the number of rounds the i-th person wants to play.
OutputIn a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at leastai rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the%I64d specifier.
Sample test(s)input33 2 2output
4input
42 2 2 2output
3Note
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times:http://en.wikipedia.org/wiki/Mafia_(party_game).
二分枚举就可以了,比赛是竟没想到用二分,二啊
#include <iostream>#include <stdio.h>#include <math.h>using namespace std;#define M 100050#define inf (((__int64)1<<62)-1)__int64 pri[M];int n;int fun(__int64 x){ int i;__int64 sum=0; for(i=0;i<n;i++){ __int64 temp=x-pri[i]; if(temp<0)return 0; sum+=temp; if(sum>inf)return 1; } if(sum>=x)return 1; return 0;}int main(){ int i; while(scanf("%d",&n)!=EOF){ for( i=0;i<n;i++){ scanf("%I64d",&pri[i]); } __int64 l=1,r=inf-1,mid,last; while(l<=r){ mid=(l+r)>>1; if(fun(mid)) r=mid-1,last=mid; else l=mid+1; } printf("%I64d\n",last); } return 0;}