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UVa 756 Biorhythms
方法:中国剩余定理
列出方程,发现是求解线性模方程,而且三个mod 两两互质,所以用中国剩余定理即可。 注意,最后求具体日期时,我采用了试一试的方法,因为方程组的解 D 和 d 之间的大小关系不确定。
code:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include <cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> #include <fstream> #include <cassert> #include <unordered_map> #include <cmath> #include <sstream> #include <time.h> #include <complex> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a)) #define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a)) #define DFOR(a,b,c) for (int (a)=(b);(a)>=(c);--(a)) #define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a)) #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) #define FOREACH(a,b) for (auto &(a) : (b)) #define rep(i,n) FOR(i,0,n) #define repn(i,n) FORN(i,1,n) #define drep(i,n) DFOR(i,n-1,0) #define drepn(i,n) DFOR(i,n,1) #define MAX(a,b) a = Max(a,b) #define MIN(a,b) a = Min(a,b) #define SQR(x) ((LL)(x) * (x)) #define Reset(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define mp make_pair #define pb push_back #define all(v) v.begin(),v.end() #define ALLA(arr,sz) arr,arr+sz #define SIZE(v) (int)v.size() #define SORT(v) sort(all(v)) #define REVERSE(v) reverse(ALL(v)) #define SORTA(arr,sz) sort(ALLA(arr,sz)) #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) #define PERMUTE next_permutation #define TC(t) while(t--) #define forever for(;;) #define PINF 1000000000000 #define newline ‘\n‘ #define test if(1)if(0)cerr using namespace std; using namespace std; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<int,int> ii; typedef pair<double,double> dd; typedef pair<char,char> cc; typedef vector<ii> vii; typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll> l4; const double pi = acos(-1.0); void gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { d = a; x = 1; y = 0; } else { gcd(b, a%b, d, y, x); y -= x*(a/b); } } ll crt(int n, ll *a, ll *m) { ll M = 1, d, y, x = 0; for (int i = 0; i < n; ++i) M *= m[i]; for (int i = 0; i < n; ++i) { ll w = M / m[i]; gcd(m[i], w, d, d, y); x = (x + y*w*a[i]) % M; } return (x+M)%M; } ll m[3] = {23, 28, 33}; ll a[3]; ll M = 23 * 28 * 33; ll d; int main() { int kase = 0; while (cin >> a[0] >> a[1] >> a[2] >> d) { if (a[0] == -1) break; ll D = crt(3, a, m); ll k =(d-D)/M; ll ans = k*M+D-d; if (ans <= 0) ans += M; //cerr << D << " " << d << " " << M << newline; cout << "Case " << ++kase << ": the next triple peak occurs in " << ans << " days." << newline; } } /* 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 */
UVa 756 Biorhythms
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