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The string APPAPT contains two PAT’s as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.
Now given any string, you are supposed to tell the number of PAT’s contained in the string.
Input Specification:
Each input file contains one test case. For each case, there is only one line giving a string of no more than 10 ^ 5 characters containing only P, A, or T.
Output Specification:
For each test case, print in one line the number of PAT’s contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.
Sample Input:
APPAPTSample Output:
2方法一：题目意思就是找出一个A左边的P个数*右边的T，构造两个数组分别存储在字符串i位置左边P的个数和右边T的个数，然后遍历字符串将合理位置的两个数组的数乘起来相加即可，不要忘了最后模1000000007,设字符串长度N ,整个复杂度就是遍历了三遍字符串所以还是O(N)的，不会超时。这种做法比较直接，也比较容易想到
AC代码
#include<iostream>#include<algorithm>#include<vector>using namespace std;#define N 1000000007typedef long long LL;string m;int main(){ LL res=0; cin >> m; vector<int> a(m.size(),0); vector<int> b(m.size(),0); for(int i=1;i<m.size();i++) if(m[i-1]=='P') a[i]=a[i-1]+1; else a[i]=a[i-1]; for(int i=m.size()-2;i>=0;i--) if(m[i+1]=='T') b[i]=b[i+1]+1; else b[i]=b[i+1]; for(int i=0;i<m.size();i++) { if(m[i]=='A') res=(res+a[i]*b[i])%N; } printf("%lld",res); return 0;}方法二：DP
#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define N 1000000007char s[100010],p[]=" PAT";int f[100010][4];int main(){ cin >> s+1; int n=strlen(s+1); f[0][0]=1; for(int i=1;i<=n;i++) for(int j=0;j<=3;j++) { f[i][j]=f[i-1][j]; if(s[i]==p[j]) f[i][j]=(f[i][j]+f[i-1][j-1])%N; } cout << f[n][3]; return 0;}转载地址：http://azbwz.baihongyu.com/