HDU-4551 生日猜猜猜 数学+枚举
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4551
水题一道。
1 //STATUS:C++_AC_15MS_232KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //define 25 #define pii pair<int,int> 26 #define mem(a,b) memset(a,b,sizeof(a)) 27 #define lson l,mid,rt<<1 28 #define rson mid+1,r,rt<<1|1 29 #define PI acos(-1.0) 30 //typedef 31 typedef __int64 LL; 32 typedef unsigned __int64 ULL; 33 //const 34 const int N=2010; 35 const int INF=0x3f3f3f3f; 36 const int MOD=100000,STA=8000010; 37 const LL LNF=1LL<<60; 38 const double EPS=1e-8; 39 const double OO=1e15; 40 const int dx[4]={-1,0,1,0}; 41 const int dy[4]={0,1,0,-1}; 42 //Daily Use ... 43 inline int sign(double x){return (x>EPS)-(x<-EPS);} 44 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 45 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 46 template<class T> inline T Min(T a,T b){return a<b?a:b;} 47 template<class T> inline T Max(T a,T b){return a>b?a:b;} 48 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 49 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 50 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 51 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 52 //End 53 54 int T,ans[N][2]; 55 int d[15]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 56 57 int main() 58 { 59 // freopen("in.txt","r",stdin); 60 int i,j,y,a,b,flag,ca=1; 61 scanf("%d",&T); 62 while(T--) 63 { 64 scanf("%d%d%d",&a,&b,&y); 65 flag= ((y%4==0 && y%100!=0) || y%400==0); 66 int k=0; 67 for(i=1;i<=12;i++){ 68 for(j=1;j<=31;j++){ 69 if(gcd(i,j)==a && lcm(i,j)==b){ 70 ans[k][0]=i; 71 ans[k++][1]=j; 72 } 73 } 74 } 75 int cnt=0,w; 76 for(i=0;i<k;i++){ 77 if(ans[i][0]!=2 && ans[i][1]<=d[ans[i][0]]){w=i;cnt++;} 78 else if(flag && ans[i][1]<=29){w=i;cnt++;} 79 else if(ans[i][1]<29){w=i;cnt++;} 80 } 81 82 printf("Case #%d: ",ca++); 83 if(cnt==0)printf("-1\n"); 84 else if(cnt==1)printf("%d/%02d/%02d\n",y,ans[w][0],ans[w][1]); 85 else printf("1\n"); 86 87 } 88 return 0; 89 }
