SRM 450 div1(practice)

250pt:

      對普通nim遊戲的改編，取石子的時候只能從前往後取，每次取非空的石子，不能取者為輸。

     我是從後往前推的，注意當前數字如果>1，不管後面是必勝態還是必敗態，肯定能調節成必勝態，如果=1，就身不由己了。。

500pt

  一開始，你有n個工廠，k個專家，每工作一輪，能獲得n*k的gold，購買一個工廠或者一個專家需要price的gold，你想要獲得的gold是target，當你具備足夠的錢買一個專家或者工廠的時候，你可以買任意多個專家或工廠。問你最小需要幾輪才能獲得target的gold。

 這是一個 看似難寫，實則更難寫的類比題。。

  注意到n k肯定是越接近越好，還有當前如果要買一個專家或者工廠，那肯定是花所有的錢去買(n,k肯定是越早增大越好)。因為下一輪就賺回來了。還有就是假設當前的錢>=price了，就需要先判斷一下是不是要買工廠或者專家了，因為有可能憑藉當前的nk，很快就能到達target了。

 代碼寫的比較渣，錯了很多遍

#include <cstdio>#include <cstring>#include <cctype>#include <cstdlib>#include <cmath>#include <ctime>#include <iostream>#include <map>#include <set>#include <list>#include <sstream>#include <queue>#include <deque>#include <stack>#include <vector>#include <bitset>#include <algorithm>using namespace std;typedef long long lld;class StrongEconomy {public:    lld ABS(lld a)    {        if(a > 0) return a;        return -a;    }    void get(lld &n,lld &k,lld &tim) {        if(tim >= ABS(n-k) )        {            lld tn = n , tk = k;            n = (tn+tk+tim) / 2;            k = (tn+tk+tim+1) / 2;        }        else        {            if(n>k) k+=tim; else n+=tim;        }    }    lld gao(lld n,lld k,lld price,lld target,lld earn) {        lld tim = earn / price;        earn -= tim * price;        get(n,k,tim);        lld tmp = mul(n,k);        if(tmp == -1) return tmp;        lld ans = (target - earn)/tmp;        if((target-earn)%tmp) ans ++;        return ans;    }    lld mul(lld a,lld b) {        lld ans = 0;        bool flag = false;        for(int i=0; i<50; i++) {            // cout<<ans<<endl;            if(b&(1LL<<i)) {                if(flag) return -1;                ans = ans + a;                if(ans > 1000000000000LL) return -1;            }            a = a + a;            if(a > 1000000000000LL) flag = true;        }        return ans;    }    lld earn(lld n, lld k, lld price, lld target)  {        if(mul(n,k) == -1) return 1;        lld round = 0;        lld earn = 0;        while(true) {            lld tmp = mul(n,k);  if(tmp == -1) return round ;            //  cout<<earn<<endl;            if(earn >= target) break;            if(earn >= price) {                lld tmp = mul(n,k);                lld tmp1 = (target - earn) / tmp ;                if((target-earn)%tmp) tmp1++;                lld tmp2 = gao(n,k,price,target,earn);            //    cout<<tmp1<<" "<<tmp2<<endl;                if(tmp1 < tmp2 ) {                    round += tmp1;                    break;                } else {                    lld tim = earn / price;                    if(tmp2 == -1) return round + 1;                    earn -= tim * price;                  //  cout<<"pre"<<" "<<n<<" "<<k<<" "<<tim<<endl;                    get(n,k,tim);                   // cout<<n<<" "<<k<<" "<<earn<<endl;                }            }            else if(price < target){                lld tmp = n * k;                lld t = (price-earn)/tmp;                if((price-earn)%tmp) t++;                round += t;                earn += t * n * k;                cout<<"earn"<<" "<<earn<<endl;            } else {                lld tmp = n * k;                lld t = (target-earn)/tmp;                if((target-earn)%tmp) t++;                round += t;                earn += t * n * k;            }          //  cout<<n<<" "<<k<<endl;        }        return round;    }// BEGIN CUT HEREpublic:    void run_test(int Case) {        if ((Case == -1) || (Case == 1)) test_case_1();    }private:    template <typename T> string print_array(const vector<T> &V) {        ostringstream os;        os << "{ ";        for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\",";        os << " }";        return os.str();    }    void verify_case(int Case, const lld &Expected, const lld &Received) {        cerr << "Test Case #" << Case << "...";        if (Expected == Received) cerr << "PASSED" << endl;        else {            cerr << "FAILED" << endl;            cerr << "\tExpected: \"" << Expected << '\"' << endl;            cerr << "\tReceived: \"" << Received << '\"' << endl;        }    }    //void test_case_0() { lld Arg0 = 4294967297LL; lld Arg1 = 4294967297LL; lld Arg2 = 1000000000000LL; lld Arg3 = 100000000000LL; lld Arg4 = 1LL; verify_case(0, Arg4, earn(Arg0, Arg1, Arg2, Arg3)); }    void test_case_1() {        lld Arg0 = 3LL;        lld Arg1 = 2LL;        lld Arg2 = 1000000000000LL;        lld Arg3 = 222LL;        lld Arg4 = 37LL;        verify_case(1, Arg4, earn(Arg0, Arg1, Arg2, Arg3));    }//void test_case_2() { lld Arg0 = 1LL; lld Arg1 = 1LL; lld Arg2 = 500000LL; lld Arg3 = 1000002LL; lld Arg4 = 1000001LL; verify_case(2, Arg4, earn(Arg0, Arg1, Arg2, Arg3)); }    //void test_case_3() { lld Arg0 = 5LL; lld Arg1 = 4LL; lld Arg2 = 15LL; lld Arg3 = 100LL; lld Arg4 = 5LL; verify_case(3, Arg4, earn(Arg0, Arg1, Arg2, Arg3)); }// END CUT HERE};// BEGIN CUT HEREint main() {    StrongEconomy ___test;    ___test.run_test(-1);}// END CUT HERE