思路:
- 题意:一个分成4种情况:0,有料就可钓一只鱼;1,可自制一包料;2,可直接抓一条鱼;3,可选择抓一条鱼或制一包料。问最多能得到多少鱼。
- 对于0,有饵就捕鱼;对于2 或 3,直接选择捕鱼,不浪费鱼饵;对于 1,默认制饵,最后加上剩下的鱼饵数除 2 就是答案(即一半用来制饵,一半用来捕鱼)。
代码实现:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n;
string s;
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n >> s;
int fish = 0, bait = 0;
for(int i = 0; i < n; i ++){
if(s[i] == '0' && bait) bait --, fish ++;
if(s[i] == '1') bait ++;
if(s[i] == '2' || s[i] == '3') fish ++;
}
cout << fish + bait / 2 << endl;
}
return 0;
}