第五题

抽卡游戏，n张普通卡，m张ssr，抽到ssr花费2金币不放回，抽到普通卡1金币放回，求取完所有ssr的期望金币数

 1 #include <iostream>
 2 using namespace std;
 3 double ans;
 4 int main()
 5 {
 6     int n, m; cin >> n >> m;
 7     for(int i=1; i<=m; i++) 
 8         ans += 2.0 + double(n) / double(i);
 9     printf("%.2lf\n", ans);
10     return 0;
11 }

第四题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int N = 1e5 + 7;
 5 LL dp[N][4];
 6 int n;
 7 int main()
 8 {
 9     cin >> n;
10     string str;
11     cin >> str;
12     for(int i=1; i<=n; i++) {
13         for(int j=0; j<4; j++) dp[i][j] = dp[i-1][j];
14 
15         if(str[i-1] == 'S') dp[i][0] += 1;
16         else if(str[i-1] == 'T') {
17             if(i>=2) dp[i][1] += dp[i-2][0];
18         }
19         else if(str[i-1] == 'A') {
20             if(i>=2) dp[i][2] += dp[i-2][1];
21         }
22         else if(str[i-1] == 'R') {
23             if(i>=2) dp[i][3] += dp[i-2][2];
24         }
25 
26     }
27     cout << dp[n][3] << endl;
28     return 0;
29 }

第三题  dfs爆栈 bfs

#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 7;
struct node {
    int x, y;
};
int dx[] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[] = {-2, -1, 1, 2, 2, 1, -1, -2};
int n, m;
string mp[N];
bool vis[N][N];
queue<node>
int ans;
bool isok(int x, int y) {
    return x>=0 && x<n && y>=0 && y<m;
}
void bfs(int x, int y) {

    node start = {x, y};
    q.push(start); vis[x][y] = 1;
    ans += 1;

    while(!q.empty()) {
        node now = q.front(); q.pop();
        x = now.x;
        y = now.y;
        for(int i=0; i<8; i++) {
            int tx = x + dx[i];
            int ty = y + dy[i];
            if(isok(tx, ty) && mp[x][y] != mp[tx][ty] && !vis[tx][ty]) {
                node tmp = {tx, ty};
                q.push(tmp); vis[tx][ty] = 1;
                ans += 1;
            }
        } 
    }
}
int main()
{
    cin >> n >> m;
    for(int i=0; i<n; i++)
        cin >> mp[i];
    int x, y;
    cin >> x >> y;
    dfs(x-1, y-1);
    cout << ans << endl;
    return 0;
}

　　

第二题 二分

x^k = y

x + y = b

(k>0, b>0) 求两条曲线在第一象限的相交面积 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int k, b;
 4 double binary_search(int k, int b) {
 5     double l = 0, r = b;
 6     while(l<=r) {
 7         double mid = (l+r) * 0.5;
 8         if(k * log(mid) > log(1e6)) {
 9             r = mid;
10             continue;
11         }
12         if(abs(mid + pow(mid, k)  - b) < 1e-9) return mid;
13         else if(mid + pow(mid, k) > b)  r = mid;
14         else                            l = mid;
15     }
16     return l;
17 }
18 int main()
19 {
20     cin >> k >> b;
21     double x = binary_search(k, b);
22     // printf("%lf %lf\n", x, x + pow(x, k));
23     double area = pow(x, k+1) / (k+1) + pow(x, k) * pow(x, k) * 0.5;
24     printf("%.6lf\n", area);
25     return 0;
26 }

 

第一题 二分

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
struct node {
	double val;
	int x, y;
	bool operator<(const node& a) const {
		return val < a.val;
	}
};
LL w[507];
vector<node> num;
int n, m;
int binary_search(LL x) {
	double key = log(x);
	int l = 0, r = num.size() - 1;
	while(l<=r) {
		int mid = (l+r) / 2;
		if(abs(num[mid].val - key) <= 1e-9) return mid;
		else if(num[mid].val > key) r = mid - 1;
		else                        l = mid + 1;
	}
	return -1;
}
int main()
{
	cin >> n >> m;
	for(int i=0; i<n; i++) cin >> w[i];
	for(int i=0; i<n; i++)
		for(int j=i+1; j<n; j++) {
			double val = log(w[i]) * w[j];
			node tmp = {val, i, j};
			num.push_back(tmp); 

			val = log(w[j]) * w[i];
			node ttmp = {val, j, i};
			num.push_back(ttmp); 
		}
	sort(num.begin(), num.end());
	for(int i=0; i<m; i++) {
		LL x; cin >> x;
		int pos = binary_search(x);
		if(pos == -1) cout << "-1 -1\n";
		else 		  cout << w[num[pos].x] << " " << w[num[pos].y] << endl; 
	}
	return 0;
}