给你一个字符串，连续的子串需要形容为kx，如BBB形容为3B，如果k为1，则输出时忽略k
通过双指针枚举即可，简单水题
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <string>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long int ll;
typedef __int64 bi;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){
return x&(-x);}
ll gcd(ll a,ll b){
return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
if(!b){
d=a,x=1,y=0;}else{
ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){
ll res=1;a%=MOD;while(b>0){
if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){
return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){
ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){
ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {
if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = {
{
1,0}, {
-1,0},{
0,1},{
0,-1} };
signed main(void)
{
int t;
cin>>t;
string s;
while(t--){
cin>>s;
int l = 0,r = 0,n = s.length();
map<char,int> m;
while(l<=r && r<n){
++m[s[l]];
while(r+1<n && s[r+1] == s[l]){
++m[s[l]];
++r;
}
if(m[s[l]] > 1) cout<<m[s[l]]<<s[l];
else cout<<s[l];
m[s[l]] = 0;
l = ++r;
}
cout<<endl;
}
}