A. QAQ
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
illustration by 猫屋 https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don‘t have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 ≤ n ≤ 100). It‘s guaranteed that the string only contains uppercase English letters.
Output
Print a single integer — the number of subsequences "QAQ" in the string.
Examples
input
Copy
QAQAQYSYIOIWIN
output
Copy
4
input
Copy
QAQQQZZYNOIWIN
output
Copy
3
题意
给一个字符串，找一共有里面多少个“QAQ”子序列（可以不连续）。
分析
可以暴力求解，这里用练习一下dp用dp解。一维不行就二维来讨论情况。
dp[i][1]表示，前i个中“Q”这种情况个数。
dp[i][2]表示，前i个中“QA”这种情况个数。
dp[i][3]表示，前i个中“QAQ”这种情况的个数。
再讨论一下当前字符是Q还是A状态转移就行了。
/// author:Kissheart ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e6+10;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
string s;
int n,dp[105][3];
int main()
{
cin>>s;
n=s.size();
if(s[0]==‘Q‘) dp[0][1]=1;
for(int i=1;i<n;i++)
{
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
if(s[i]==‘A‘)
dp[i][2]+=dp[i][1];
else if(s[i]==‘Q‘)
{
dp[i][1]++;
dp[i][3]+=dp[i][2];
}
}
printf("%d\n",dp[n-1][3]);
return 0;
}
原文地址：https://www.cnblogs.com/Kissheart/p/10390193.html