# [POJ 1006] Biorhythms【中国剩余定理】

## Problem:

Time Limit: 1000MS |
Memory Limit: 10000K |

Description

Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.

Input

Output

For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form:

Case 1: the next triple peak occurs in 1234 days.

Use the plural form ``days'' even if the answer is 1.

Sample Input

0 0 0 0 0 0 0 100 5 20 34 325 4 5 6 7 283 102 23 320 203 301 203 40 -1 -1 -1 -1

Sample Output

Case 1: the next triple peak occurs in 21252 days. Case 2: the next triple peak occurs in 21152 days. Case 3: the next triple peak occurs in 19575 days. Case 4: the next triple peak occurs in 16994 days. Case 5: the next triple peak occurs in 8910 days. Case 6: the next triple peak occurs in 10789 days.

Source

## Solution:

本题是**中国剩余定理**的水题。

求解同余方程组

*x ≡ a*_{1}(mod m_{1})*x ≡ a*_{2}(mod m_{2})*……**x ≡ a*_{n}(mod m_{n})

首先构造 *M = Π _{1≤i≤n} m_{i}*，令

*M*，t

_{i}= M / m_{i}_{i}为 M

_{i}关于 m

_{i}的乘法逆元，则有

*x ≡ ∑*_{1≤i≤n}a_{i }M_{i }t_{i}(mod M)

Code #1 是带有求逆元的代码。而考虑到本题 m_{i} 均已给出，可以简化为 Code #2。

## Code #1: O(T+logC), C为1e3级别的常数 [668K, 94MS]

#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; int cas = 0; ll phy, emo, inter, dat; inline ll exgcd(ll a, ll b, ll &x, ll &y){ if(!b) {x = 1, y = 0; return a;} ll d = exgcd(b, a % b, y, x); y -= a / b * x; return d; } inline ll inv(ll a, ll Mod){ ll x, y; exgcd(a, Mod, x, y); return (x % Mod + Mod) % Mod; } int main(){ ll k1 = 28 * 33 * inv(28 * 33, 23); ll k2 = 23 * 33 * inv(23 * 33, 28); ll k3 = 23 * 28 * inv(23 * 28, 33); while(scanf("%lld%lld%lld%lld", &phy, &emo, &inter, &dat) != EOF && dat != -1){ ll ans = (phy * k1 + emo * k2 + inter * k3 - dat + 21251) % 21252 + 1; printf("Case %d: the next triple peak occurs in %lld days.\n", ++cas, ans); } return 0; }

## Code #2: O(T) [660K, 47MS]

#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; int cas = 0; ll phy, emo, inter, dat; int main(){ while(scanf("%lld%lld%lld%lld", &phy, &emo, &inter, &dat) != EOF && dat != -1){ ll ans = (phy * 5544 + emo * 14421 + inter * 1288 - dat + 21251) % 21252 + 1; printf("Case %d: the next triple peak occurs in %lld days.\n", ++cas, ans); } return 0; }

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