C. Kuroni and Impossible Calculation(鸽巢定理) 题意: 求∏ a b s ( a [ i ] − a [ j ] ) m o d m m < = 1000 n < = 2 e 5 \prod abs(a[i]-a[j])~mod~m\\ m<=1000 n<=2e5 ∏abs(a[i]−a[j]) mod mm<=1000n<=2e5 思路: 完全没想到这是个**题,运用鸽巢定...
https://codeforces.com/problemset/problem/1305/C 题目大意:给出NM,代表N个数字,mod为M。 序列中两两求差值,求将所有差值相乘,最后模M的值。 思考过程: 因为如果n大于m的情况,总会有两个数字的模是相同的,这样就会导致最后的结果为0,所以,只需要考虑n小于等于m的情况即可,而m又很小,就可以用暴力来做这个...
#include <bits/stdc++.h> #define int long long using namespace std; int32_t main(){ int n,m;cin>>n>>m; vector<int> arrmv(n); for(int i=0;i<n;i++)cin>>arrmv[i]; if(n>m)cout<<0<<endl; else{ int ans=1; for(int i=0;i<n;i++) for(int j=i+1;j<n;j++){...
n≤m时,直接枚举i和j计算答案即可。 n>m i,j(i≠j) ai≡bj(modm) ai−bj≡0(modm) ∏1≤i≤j≤n|ai−aj|≡0(modm) #include<bits/stdc++.h>usingnamespacestd;typedeflonglongll; ll n,m,a[200000+10];intmain(){ cin >> n >> m;for(ll i=1;i<=n;i++)cin >> a[i];if...
第一反应就是ο(n2)ο(n2)的暴力枚举,由于n范围过大而放弃,同时注意到m的范围很小,mod(m)的余数不会超过m种,那么由鸽巢原理可得前m+1个数里面必定有至少两个数关于mod(m)同余,那么这两个数mod(m)的差值必定为0,即答案为0。对于剩下的n<m的情况直接ο(n2)ο(n2)的枚举就可以了 代码: #include<bi...
C_Kuroni_and_Impossible_Calculation.cpp C_Largest_Subsequence.cpp C_Lazy_Narek.cpp C_Least_Prefix_Sum.cpp C_Least_Prefix_Sum.exe C_Left_and_Right_Houses.cpp C_Light_Switches.cpp C_Lining_Up_2.cpp C_Little_Girl_and_Maximum_Sum.cpp C_Long_Multiplication.cpp C_Loong_Tracking.cpp C_Lucky...
In the second test case, f(x) is x+2 and g(x) is x+3, their product h(x) being x2+5x+6, so the answer can be any of the powers as no coefficient is divisible by the given prime. 题意: 给定两个多项式长度 n 和 m ,再给定每一项的系数,由常数项到最高次项排序,其中每个多项式...
ACM思维题训练集合 A bracket sequence is a string containing only characters “(” and “)”. 风骨散人Chiam 2020/10/29 3750 Codeforce-Ozon Tech Challenge 2020-C. Kuroni and Impossible Calculation(鸽笼原理) inputlinemodulooutputsample To become the king of Codeforces, Kuroni has to solve the ...
C_Kuroni_and_Impossible_Calculation.cpp C_Largest_Subsequence.cpp C_Least_Prefix_Sum.cpp C_Least_Prefix_Sum.exe C_Left_and_Right_Houses.cpp C_Light_Switches.cpp C_Lining_Up_2.cpp C_Little_Girl_and_Maximum_Sum.cpp C_Long_Multiplication.cpp C_Loong_Tracking.cpp C_Lucky_Numbers.cpp C_Luo...
printf("%d", ans.top() +1); ans.pop(); } }return0; } C.Kuroni and Impossible Calculation 被C卡了半天(丢人),可以发现模很小,那么当n>m的时候,一定会存在ai 和 aj 使ai % m == aj % m,那这两个数相减就是0了,所以n>m,答案一定是0,那1000一下直接暴力就好了 ...