CTU Open 2008（未完工）

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链接：CTU Open 2008

【2014/05/15】今晚做了一下CTU 2008的这套题，最后的rank是3道题。基本是水题啊，我们做的是4个小时，如果完整做，我想应该还会出掉D题，主要是D题很繁琐，weikd写起都说烦。

A - Alea iacta est

B - On-Line Banking

【题意】纯模拟题，模拟银行存钱，取钱，转钱的操作。注意下细节~

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <string>#define EPS 1e-8using namespace std;char str[1005];char cmd[1005];char name[1005],name1[1005];map<string,double> ac;void create(string id){    printf("create: ");    if (ac.find(id)!=ac.end())    {        printf("already exists\n");    }    else    {        ac[id] = 0;        printf("ok\n");    }}void deposit(string id, double x){    printf("deposit %.2f: ",x);    if (ac.find(id)==ac.end())    {        printf("no such account\n");    }    else    {        ac[id] = ac[id]+x;        printf("ok\n");    }}void withdraw(string id,double x){    printf("withdraw %.2f: ",x);    if (ac.find(id)==ac.end())    {        printf("no such account\n");    }    else    {        if (ac[id]+EPS<x)        {            printf("insufficient funds\n");        }        else        {            ac[id]=ac[id]-x;            printf("ok\n");        }    }}void transfer(string ida,string idb,double x){    printf("transfer %.2f: ",x);    if (ac.find(ida)==ac.end()||ac.find(idb)==ac.end())    {        printf("no such account\n");    }    else    {        if (ida==idb)        {            printf("same account\n");        }        else        {            if (ac[ida]+EPS<x)            {                printf("insufficient funds\n");            }            else            {                ac[ida]=ac[ida]-x;                ac[idb]=ac[idb]+x;                char ia = *ida.rbegin();                char ib = *idb.rbegin();                if (ia==ib)                {                    printf("ok\n");                }                else                {                    printf("interbank\n");                }            }        }    }}int main(){    #ifdef HotWhite    freopen("in.txt", "r", stdin);    #endif    int n;    double mn;    while(scanf("%d",&n))    {        if (!n) break;        ac.clear();        for (int i = 0; i < n; i++)        {            scanf("%s %lf",str,&mn);            ac[str]=mn;        }        string s1,s2;        while(scanf("%s",cmd))        {            if (strcmp(cmd,"end")==0) break;            if (strcmp(cmd,"create")==0)            {                scanf("%s",name);                s1=name;                create(s1);            }            if (strcmp(cmd,"deposit")==0)            {                scanf("%s%lf",name,&mn);                s1=name;                deposit(s1,mn);            }            if (strcmp(cmd,"withdraw")==0)            {                scanf("%s%lf",name,&mn);                s1=name;                withdraw(s1,mn);            }            if (strcmp(cmd,"transfer")==0)            {                scanf("%s%s%lf",name,name1,&mn);                s1=name;                s2=name1;                transfer(s1,s2,mn);            }        }        printf("end\n\n");    }    printf("goodbye\n");    return 0;}

C - International Collegiate Programming Contest

D - Careful Declaration

E - Stock Exchange

【题意】每一组case给你一些买家的姓名和最高出价，还有卖家的姓名和最低售价。然后计算出所有成交信息。
【思路】水题，直接暴力模拟，比较价钱即可。代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAX = 1010;struct BID{    char name[25], type[5];    double price;};BID bid[MAX], buy[MAX], sell[MAX];int main() {#ifdef HotWhite    freopen("in.txt", "r", stdin);#endif    int N, flag;    char issuer[12];    while(scanf("%d%s", &N, issuer) != EOF && N)    {        int cntb = 0, cnts = 0;        for(int i=0; i<N; ++i)        {            scanf("%s%s%lf", bid[i].name, bid[i].type, &bid[i].price);            if(strcmp(bid[i].type, "buy") == 0)            {                buy[cntb++] = bid[i];            }            if(strcmp(bid[i].type, "sell") == 0)            {                sell[cnts++] = bid[i];            }        }        printf("%s\n", issuer);        for(int i=0; i<N; ++i)        {            flag = 0;            printf("%s:", bid[i].name);            if(strcmp(bid[i].type, "buy") == 0)            {                for(int j=0; j<cnts; ++j)                {                    if(sell[j].price <= bid[i].price)                    {                        printf(" %s", sell[j].name);                        flag = 1;                    }                }            }            if(strcmp(bid[i].type, "sell") == 0)            {                for(int j=0; j<cntb; ++j)                {                    if(buy[j].price >= bid[i].price)                    {                        printf(" %s", buy[j].name);                        flag = 1;                    }                }            }            if(!flag)                printf(" NO-ONE");            printf("\n");        }    }    return 0;}

F - Wooden Fence

G - Safe Gambling

H - Government Help

【题意】经济危机，政府给予bankA和bankB补助，补助是按照一包一包的算。总共有N包，为了公平起见，怎么分配才能使得最后bankA和bankB得到的补助之差最小。

【思路】贪心策略，从小到大排序，先把最小的给A。然后从最大的开始先给B，再给A，依次分配下去，直到分完。代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[50005];int main() {#ifdef HotWhite   // freopen("in.txt", "r", stdin);#endif    int n;    while(scanf("%d",&n))    {        if (!n) break;        for (int i = 0; i < n; i++)        {            scanf("%d",&a[i]);        }        sort(a,a+n);        printf("%d-A",a[0]);        int k=0;        for (int i = n-1; i >=1; i--,k++)        {            printf(" %d-%c",a[i],k&1?'A':'B');        }        printf("\n");    }    return 0;}

I - Tree Insertions

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