[hdu5373 The shortest problem]模拟

http://acm.hdu.edu.cn/showproblem.php?pid=5373

思路：按题意来即可。

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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + ; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ int get(int sum, int &x, int &mark) { if (sum == ) { x = ; mark = ; return ; } mark = ; x = ; int ans = ; while (sum) { mark ^= ; if (mark) ans += sum % ; else ans -= sum % ; x += sum % ; sum /= ; } return ans; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int n, t, cas = ; while (cin >> n >> t, n != - && t != -) { int mark, sum, ans; ans = get(n, sum, mark); for (int i = ; i < t; i ++) { int x, buf = get(sum, x, mark); if (mark) ans = -ans + buf; else ans = ans + buf; sum += x; } printf("Case #%d: %s\n", ++ cas, ans % ? "No" : "Yes"); } return ; }