Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)
二进制专场,还被FST一题。 A、签到,题意是,给定a,b,c f(0)=a,f(1)=b,f(n)=f(n-1)^f(n-2) 求f(n) 稍微化简就看得出f(x)是按照a,b,a^b这样在循环出现 因为a^b^a=b,a^b^b=a 代码: #include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair