PAT训练（4）

1092. To Buy or Not to Buy (20)

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=75;

int main()
{
string exist,want;
int table[maxn],remain,mark[maxn];
while(cin>>exist>>want)
{
int i;
memset(table,0,sizeof(table));
memset(mark,0,sizeof(mark));
for(int i=0;i<exist.size();i++)
table[exist[i]-'0']++;
for(i=0;i<want.size();i++)
table[want[i]-'0']--;

remain=0;
for(i=0;i<maxn;i++)
if(table[i]>=0)
remain+=table[i];
else
{
remain=table[i];
i++;
break;
}
for(;i<maxn;i++)
if(table[i]<0)
remain+=table[i];

if(remain<0)
cout<<"No "<<-remain<<endl;
else
cout<<"Yes "<<remain<<endl;
}
}

1093. Count PAT’s (25)

#include<iostream>
#include<cstring>
using namespace std;
const long long mod=1000000007;
const int maxn=100005;

int main()
{
//找到'A'前的'P'个数和其后的'T'个数,二者相乘,O(3n).
string x;
long long ans;
int num[2][maxn],np,nt;
while(cin>>x)
{
memset(num,0,sizeof(num));
np=nt=ans=0;
for(int i=0;i<x.size();i++)
{
if(x[i]=='P')
np++;
num[0][i]=np;
}
for(int i=x.size()-1;i>=0;i--)
{
if(x[i]=='T')
nt++;
num[1][i]=nt;
}

for(int i=0;i<x.size();i++)
if(x[i]=='A')
ans+=num[0][i]*num[1][i];

cout<<ans%mod<<endl;
}
}