codeforces 349C Mafia [贪心]/[二分答案]
2017-01-10 02:05
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题目链接:http://codeforces.com/contest/349/problem/C
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C. Mafia
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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?
Input
The 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.
Output
In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai 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.
Examples
input
3
3 2 2
output
4
input
4
2 2 2 2
output
3
Note
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).
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题目大意:就是有n个人做游戏,每次需要有一个人作为”Mafia”,剩下的人作为“player”,现在有n个人每个人想做“player”的次数,问你满足所有人想做“player”的次数意愿的同时最小游戏次数是多少。
题目大意:
我是用二分答案做的,然后看了下别人的 发现直接贪心就可以了。。
对结果进行二分,然后check的时候,只要判断所有人能做”Mafia”的次数能不能满足mid就行了。
但是神TM二分范围啊 ,本来想找找自信,结果r的范围TM的WA三发、、、
首先l一定是max{ai}
然后r的范围我设1ll<<62居然不行。1ll<<60居然还TM不行。?
最后1e12才过、、
附本题代码
—————————————————–。
———————————————————————————————————————–.
C. Mafia
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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?
Input
The 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.
Output
In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai 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.
Examples
input
3
3 2 2
output
4
input
4
2 2 2 2
output
3
Note
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).
———————————————————————————————————————–.
题目大意:就是有n个人做游戏,每次需要有一个人作为”Mafia”,剩下的人作为“player”,现在有n个人每个人想做“player”的次数,问你满足所有人想做“player”的次数意愿的同时最小游戏次数是多少。
题目大意:
我是用二分答案做的,然后看了下别人的 发现直接贪心就可以了。。
对结果进行二分,然后check的时候,只要判断所有人能做”Mafia”的次数能不能满足mid就行了。
但是神TM二分范围啊 ,本来想找找自信,结果r的范围TM的WA三发、、、
首先l一定是max{ai}
然后r的范围我设1ll<<62居然不行。1ll<<60居然还TM不行。?
最后1e12才过、、
附本题代码
—————————————————–。
#include <bits/stdc++.h> using namespace std; #define INF (~(1<<31)) #define INFLL (~(1ll<<63)) #define pb push_back #define mp make_pair #define abs(a) ((a)>0?(a):-(a)) #define lalal puts("*******"); #define s1(x) scanf("%d",&x) #define Rep(a,b,c) for(int a=(b);a<=(c);a++) #define Per(a,b,c) for(int a=(b);a>=(c);a--) typedef long long int LL ; typedef unsigned long long int uLL ; const int MOD = 1e9+7; const int N = 100000+8; const double eps = 1e-6; const double PI = acos(-1.0); void fre(){ freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); } /*************************************************/ LL a ; int n; bool check(LL x){ LL res = 0; Rep(i,1,n) res+=x-a[i]; if(res>=x) return true; return false; } int main(){ while(~s1(n)){ LL l = 0,r=1e12,mid,ans=-1; Rep(i,1,n) scanf("%I64d",&a[i]),l=max(l,a[i]); while(l<=r){ mid=(l+r)>>1; if(check(mid)){ ans = mid; r=mid-1; } else l=mid+1; } printf("%I64d\n",ans); } return 0; }
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