题意:给你两个闭区间【a,b】,【c,d】,分别从中等可能的跳出 x 和 y ,求(x+y)%p == m的概率
分析:
假如是【3,5】 【4,7】 p = 2 , m = 1;
则所有的和
7 8
9 10
8 9
10 11
9 10
11 12
1 2
3 3 2
1
后面一行出现次数,可以发现可以分成三部分,第一部分递增的等差数列,第二部分值都相等,第三部分等差数列
然后用等差数列求和公式求和就ok
AC代码:
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> #include<stack> #define MAXN 100005 using namespace std; typedef long long LL; LL p,m; LL gcd(LL a,LL b) { if(b==0) return a; return gcd(b,a%b); } int main() { //freopen("Input.txt","r",stdin); int T; scanf("%d",&T); for(int cas = 1; cas<=T; cas++) { LL a,b,c,d,ans = 0; scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m); if((b-a)>(d-c)) swap(a,c),swap(b,d); long long t1 = (a+c)%p; long long add = (m - t1 + p)%p; long long cnt1 = (a+c + add-m)/p; long long t2 = (b+c-1)%p; long long sub = (t2 - m + p)%p; long long cnt2 = (b+c-1-sub-m)/p; //cout<<t2<<" "<<sub<<endl; ans += (cnt2 - cnt1 + 1)*(1+add) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2 * p; t1 = (b+c)%p; add = (m - t1 + p)%p; cnt1 = (b+c+add-m)/p; t2 = (a+d)%p; sub = (t2 - m + p)%p; cnt2 = (a+d-sub-m)/p; ans += (cnt2 - cnt1 + 1)*(b-a+1); t1 = (a+d+1)%p; add = (m - t1 + p)%p; cnt1 = (a+d+1+add-m)/p; t2 = (b+d)%p; sub = (t2 - m + p)%p; cnt2 = (b+d-sub-m)/p; ans += (cnt2 - cnt1 + 1)*(1+sub) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2*p; long long tot = (b-a+1)*(d-c+1); long long GCD = gcd(ans,tot); ans /= GCD; tot /= GCD; printf("Case #%d: %I64d/%I64d\n",cas,ans,tot); } return 0; }
时间: 2024-10-04 11:53:23