A - Anton and Polyhedrons 【模拟】
简单题。map<string,int>mp累计字符串权值
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
|
#include <bits/stdc++.h>#define scf0(a) scanf("%s", &a)#define scf1(a) scanf("%d", &a)#define scf2(a, b) scanf("%d%d", &a, &b)#define scf3(a, b, c) scanf("%d%d%d", &a, &b, &c)#define scf4(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)#define MEM(a, b) memset(a, b, sizeof(a))#define PII pair<int, int>using namespace std;int main() { map<string,int>mp; mp["Tetrahedron"] = 4; mp["Cube"] = 6; mp["Octahedron"] = 8; mp["Dodecahedron"] = 12; mp["Icosahedron"] = 20; string op; int n; cin >> n; int sum = 0; while(n--) { cin >> op; sum += mp[op]; } cout << sum << endl;} |
题意：给定两门课的不同课程安排，各选择一个，使得两门课的间隔时间最短。
解法：求两门课最小右界(minn、minm)和最大左界(maxn、maxm)。结果ans = max( 0, max(maxn-minm, maxm-minn) )
（做的时候初始化minn写成0x3f3f3f，差了一个3f...粗心啊）
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
|
#include <bits/stdc++.h>#define scf0(a) scanf("%s", &a)#define scf1(a) scanf("%d", &a)#define scf2(a, b) scanf("%d%d", &a, &b)#define scf3(a, b, c) scanf("%d%d%d", &a, &b, &c)#define scf4(a, b, c, d) scanf("%d%d%d%d", &a, &b, &c, &d)#define MEM(a, b) memset(a, b, sizeof(a))#define PII pair<int, int>using namespace std;int main() { int n, m, x, y; scf1(n); int maxn = -1, minn = 0x3f3f3f3f; for(int i = 0; i < n; i++) { scf2(x,y); maxn = max(maxn, x); minn = min(minn, y); } scf1(m); int maxm = -1, minm = 0x3f3f3f3f; for(int i = 0; i < m; i++) { scf2(x, y); maxm = max(maxm, x); minm = min(minm, y); } int ans = maxm - minn; ans = max(ans, maxn - minm); cout << ( ans > 0 ? ans : 0) << endl;} |
C - Anton and Fairy Tale 【数学题】
题意：仓库容量n，每天补充粮食m份但不能超过仓库容量，第i天麻雀吃掉i份粮食，求哪一天仓库会没有粮食
第 i 天 开始 结束
1 n n-1
2 n n-2
3 n n-3
..
m n n-m
-------------------------------
m+1 n n-m-1
m+2 n-1 n-m-3
..
m+x n-x n - m - x(x+1)/2 <= 0
一、直接解方程做法：
所以可以把时间分成两部分：1.前m天（包括第m天）2.第m天以后
解方程 n - m - x(x+1)/2 得 x = ( -1+sqrt(1+8(n-m)) ) / 2
然后！！注意精度问题。。
1.手写个sqrt避免例如sqrt(9)=2.9999向下取整得到2的情况。。
2.总之得到一个手算精确的结果，然后向上取整。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
|
#include <bits/stdc++.h>using namespace std;long long SQRT(long long x) { long long temp = sqrt(x); temp -= 5; for(temp;;temp++) { if(temp * temp == x) { return temp; } if(temp * temp > x) { return temp - 1; } }}int main() { long long n, m; cin >> n >> m; if(n <= m ) { cout << n << endl; return 0; } long long x = 1 + 8 * (n - m); long long temp = SQRT(x); //求得一个准确的sqrt if(temp * temp == x) { //如果sqrt后是整数 long long ans = -1 + temp; if(ans % 2 == 0) cout << ans / 2 + m << endl; else cout << ans / 2 + m + 1 << endl; } else { //sqrt后不是整数，那肯定要向上取整 long long ans = -1 + temp; cout << ans / 2 + m + 1<< endl; }} |
但因为sqrt精度丢失只会变小，所以也可以用直接用sqrt得出一个ans，然后judge一下ans是否满足题意（ ans*(ans+1)/2 >= n - m ）即可。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
|
#include <bits/stdc++.h>using namespace std;int main() { long long n, m; cin >> n >> m; if(n <= m ) { cout << n << endl; return 0; } long long x = ( sqrt ( 1 + 8 * (n - m) ) - 1) / 2; while(1) { long long temp = (x + 1) * x / 2; if(temp >= n - m) { cout << x + m << endl; return 0; } x++; }} |
二、二分结果+judge做法：求左界。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
|
#include <bits/stdc++.h>using namespace std;long long n, m;bool judge(unsigned long long x) { if( x * (x + 1) >= 2 * (n - m) ) return true; else return false;}int main() { cin >> n >> m; if(n <= m) { cout << n << endl; return 0; } long long low = 0LL, high = 1LL << 31; while(low < high) { long long mid = low + high >> 1; if(judge(mid)) high = mid; else low = mid+1; } cout <<high + m << endl;} |