标签:inf long code utc char gcd 题意 set mes
给出两个数A和B,要求在其中填入N个数字,使得相邻两个数字之差的绝对值在C到D之间
即要构造N个绝对值在CD之间的数其和为B-A,考虑有m个大于0,N-m个小于0,在这样的情况下的和的范围是$[Cm-(N-m)d, Dm-(N-m)C]$,枚举m依次check即可
#include <bits/stdc++.h> using namespace std; namespace my_header { #define pb push_back #define mp make_pair #define pir pair<int, int> #define vec vector<int> #define pc putchar #define clr(t) memset(t, 0, sizeof t) #define pse(t, v) memset(t, v, sizeof t) #define bl puts("") #define wn(x) wr(x), bl #define ws(x) wr(x), pc(‘ ‘) const int INF = 0x3f3f3f3f; typedef long long LL; typedef double DB; inline char gchar() { char ret = getchar(); for(; (ret == ‘\n‘ || ret == ‘\r‘ || ret == ‘ ‘) && ret != EOF; ret = getchar()); return ret; } template<class T> inline void fr(T &ret, char c = ‘ ‘, int flg = 1) { for(c = getchar(); (c < ‘0‘ || ‘9‘ < c) && c != ‘-‘; c = getchar()); if (c == ‘-‘) { flg = -1; c = getchar(); } for(ret = 0; ‘0‘ <= c && c <= ‘9‘; c = getchar()) ret = ret * 10 + c - ‘0‘; ret = ret * flg; } inline int fr() { int t; fr(t); return t; } template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); } template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); } template<class T> inline char wr(T a, int b = 10, bool p = 1) { return a < 0 ? pc(‘-‘), wr(-a, b, 0) : (a == 0 ? (p ? pc(‘0‘) : p) : (wr(a/b, b, 0), pc(‘0‘ + a % b))); } template<class T> inline void wt(T a) { wn(a); } template<class T> inline void wt(T a, T b) { ws(a), wn(b); } template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); } template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); } template<class T> inline T gcd(T a, T b) { return b == 0 ? a : gcd(b, a % b); } template<class T> inline T fpw(T b, T i, T _m, T r = 1) { for(; i; i >>= 1, b = b * b % _m) if(i & 1) r = r * b % _m; return r; } }; using namespace my_header; LL n, a, b, c, d; int main() { #ifdef lol freopen("b.in", "r", stdin); freopen("b.out", "w", stdout); #endif fr(n, a, b); fr(c, d); bool ok = false; for (int i = 0; i < n; ++i) { if (c * i - d * (n - i - 1) <= b - a && b - a <= d * i - c * (n - i - 1)) ok = true; } if (ok) { puts("YES"); } else puts("NO"); return 0; }
标签:inf long code utc char gcd 题意 set mes
原文地址:http://www.cnblogs.com/ichn/p/7580213.html