(Incomplete) UVa 138 Street Numbers
Posted 大四开始ACM
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方法:暴力
设home的序号为n,街尾序号为N,列出方程 (n-1)*n/2 = (N+n+1)*(N-n)/2, 化简得 2*n*n = (N+1)*N。枚举N再检查是否有解。可以直接求,或者打表。
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> #include <iomanip> #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); vector<ll> ans; int main() { ll i = 2; while (ans.size() < 10) { ++i; ll n = sqrt((i+1)*i/2); if (n * n == (i+1)*i/2) ans.pb(i); } for (auto i : ans) cout << setw(10) << (ll) sqrt((i+1)*i/2) << setw(10) << i << newline; }
据说可以转化成佩尔方程,待学习。
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