Codeforces Round #259 (Div. 2) 題解
A. Little Pony and Crystal Mine
time limit per test 1 second memory limit per test 256 megabytes input standard input output standard outputTwilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n × n matrix with a diamond inscribed into it.
You are given an odd integer n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
Input
The only line contains an integer n
Output a crystal of size n.
Sample test(s) input3output
*D* DDD *D*input
5output
**D** *DDD* DDDDD *DDD* **D**input
7output
***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***解題思路:
水題,推出公式,直接列印即可。
程式碼:
#include <cstdio> #include <cstring> using namespace std; int main() { int n; scanf("%d", &n); for(int i = 1; i <= n; ++i) { if(i <= n/2) { int t = (n+1-2*i)/2; for(int j = 0;j< t; ++j) printf("*"); for(int j = 0; j < 2*i-1; ++j) printf("D"); for(int j=0; j<t; ++j) printf("*"); printf("\n"); } else { int w = 2*n+1-2*i, t = (n-w)/2; for(int j = 0; j < t; ++j) printf("*"); for(int j = 0; j <w; ++j) printf("D"); for(int j = 0; j < t; ++j ) printf("*"); printf("\n"); } } return 0; }
B. Little Pony and Sort by Shift
time limit per test 1 second memory limit per test 256 megabytes input standard input output standard outputOne day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?
InputThe first line contains an integer n (2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).
OutputIf it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.
Sample test(s) input2 2 1output
1input
3 1 3 2output
-1input
2 1 2output
0題意:
給一個序列,每次可以把序列的最後一個數移到最前面,如果可以使序列遞增(嚴格來說是非遞減),輸入最少移動次數,否則,輸出-1;
解題思路:
先將序列掃一遍,遇到非遞增的位置(記為t)跳出,接著從從t+1開始掃一遍,如果,後面的序列遞增,則輸出其長度,即為答案,否則,輸出-1。因為如果要按照題中所述方式移動,使序列變為遞增,序列應當是本來就是遞增的,或者是可以分為兩個連續的遞增子序列的。
程式碼:
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 1e5 + 10;
int n, a[MAXN];
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", &a[i]);
bool flag = true;
int ans = 0, t = 0, i;
for(i = 0; i < n-1; ++i)
{
t = i;
if(a[i] > a[i+1]) break;
}
if(i != n-1)
{
for(int i = t+1; i < n; ++i)
{
ans++;
if(a[i] > a[(i+1)%n])
{
flag = false;
break;
}
}
}
if(flag)
printf("%d\n", ans);
else
printf("-1\n");
return 0;
}
C. Little Pony and Expected Maximum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the m-th face contains mdots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability . Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dice n times.
InputA single line contains two integers m and n (1 ≤ m, n ≤ 105).
OutputOutput a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4.
Sample test(s) input6 1output
3.500000000000input
6 3output
4.958333333333input
2 2output
1.750000000000解題思路: 求數學期望,公式P = m - (1/m)^n - (2/m)^n - ... -((m-1)/m)^n 程式碼:
#include <cstdio>
#include <cstring>
#include <cmath>
int main()
{
int m, n;
scanf("%d%d", &m, &n);
double ans = m;
for(int i = 1; i < m; ++i)
ans -= pow(i*1.0/m, n);
printf("%.12f\n", ans);
return 0;
}