标签：ace void 不同 begin ext algo 人生 sample map
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
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.
就是求同余方程组
$$ \left\{
\begin{aligned}
x ≡ p(\mod 23) \\
x ≡ e(\mod 28) \\
x ≡ i(\mod 33)
\end{aligned}
\right.
$$
大于$d$的最小正整数解。（喻队）孙子定理乱搞就好了。
1 //It is made by Awson on 2017.10.29 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <stack> 7 #include <queue> 8 #include <vector> 9 #include <string> 10 #include <cstdio> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Min(a, b) ((a) < (b) ? (a) : (b)) 17 #define Max(a, b) ((a) > (b) ? (a) : (b)) 18 #define Abs(x) ((x) < 0 ? (-(x)) : (x)) 19 using namespace std; 20 const int MOD = 23*28*33; 21 22 int a[5], m[5] = {0, 23, 28, 33}, n = 3, d; 23 int casecnt; 24 25 int ex_gcd(int a, int b, int &x, int &y) { 26 if (b == 0) { 27 x = 1, y = 0; return a; 28 } 29 int gcd = ex_gcd(b, a%b, x, y); 30 int t = x; 31 x = y; 32 y = t-a/b*y; 33 return gcd; 34 } 35 int inv(int a, int b) { 36 int x, y; 37 ex_gcd(a, b, x, y); 38 return (x%b+b)%b; 39 } 40 int CRT() { 41 int M = 1, ans = 0; 42 for (int i = 1; i <= n; i++) M *= m[i]; 43 for (int i = 1; i <= n; i++) 44 (ans += a[i]*(M/m[i])*inv(M/m[i], m[i])) %= M; 45 return (ans+M)%M; 46 } 47 void work() { 48 casecnt++; 49 if (d == -1) return; 50 int ans = CRT(); 51 if (ans <= d) ans += MOD-d; 52 else ans -= d; 53 printf("Case %d: the next triple peak occurs in %d days.\n", casecnt, ans); 54 } 55 int main() { 56 while (~scanf("%d%d%d%d", &a[1], &a[2], &a[3], &d)) work(); 57 return 0; 58 }
标签：ace void 不同 begin ext algo 人生 sample map
原文地址：http://www.cnblogs.com/NaVi-Awson/p/7749858.html
