Girls Love 233
Girls Love 233
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
As you see, luras sneaked into another girl's QQgroup to meet her indescribable aim.
However, luras can only speak like a cat. To hide her real identity, luras is very careful to each of her words.
She knows that many girls love saying "233",however she has already made her own word at first, so she needs to fix it.
Her words is a string of length n,and each character of the string is either '2' or '3'.
Luras has a very limited IQ which is only m.
She could swap two adjacent characters in each operation, which makes her losing 2 IQ.
Now
the question is, how many substring "233"s can she make in the string
while her IQ will not be lower than 0 after her operations?
for example, there is 1 "233" in "2333", there are 2 "233"s in "2332233", and there is no "233" in "232323".
and as for each case,
the first line are two integers n and m,which are the length of the string and the IQ of luras correspondingly.
the second line is a string which is the words luras wants to say.
It is guaranteed that——
1 <= T <= 1000
for 99% cases, 1 <= n <= 10, 0 <= m <= 20
for 100% cases, 1 <= n <= 100, 0<= m <= 100
there should be one integer in the line which represents the largest possible number of "233" of the string after her swap.
6 2
233323
6 1
233323
7 4
2223333
1
2
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+;
const int N=1e3+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p%mod;p=p*p%mod;q>>=;}return f;}
int n,m,k,t,dp[][][],pos[],cnt,cas,ret;
char a[];
int main()
{
int i,j;
scanf("%d",&cas);
while(cas--)
{
scanf("%d%d",&n,&m);
m/=;
ret=cnt=;
scanf("%s",a+);
for(i=;i<=n;i++)if(a[i]=='')pos[++cnt]=i;
pos[++cnt]=i;
memset(dp[],-,sizeof(dp[]));
dp[][][m]=;
for(i=;i<=cnt;i++)
{
for(j=i;j<=n+;j++)
{
for(k=;k<=m;k++)
{
dp[i][j][k]=-;
for(t=i-;t<j;t++)
{
if(k+abs(j-pos[i])<=m&&dp[i-][t][k+abs(j-pos[i])]!=-)
{
dp[i][j][k]=max(dp[i][j][k],dp[i-][t][k+abs(j-pos[i])]+(j-t>&&i>));
}
}
if(i==cnt&&j==n+&&ret<dp[i][j][k])ret=dp[i][j][k];
}
}
}
printf("%d\n",ret);
}
return ;
}
/*
1
5 19
33233
ans:1
*/
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