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        CodeforcesRound#259(Div.2)題解_html/css

        來(lái)源:懂視網(wǎng) 責(zé)編:小采 時(shí)間:2020-11-27 15:54:27
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        CodeforcesRound#259(Div.2)題解_html/css

        CodeforcesRound#259(Div.2)題解_html/css_WEB-ITnose:A. Little Pony and Crystal Mine time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (
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        導(dǎo)讀CodeforcesRound#259(Div.2)題解_html/css_WEB-ITnose:A. Little Pony and Crystal Mine time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (

        A. Little Pony and Crystal Mine

        time limit per test

        1 second

        memory limit per test

        256 megabytes

        input

        standard input

        output

        standard output

        Twilight 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 (3?≤?n?≤?101; n is odd).

        Output

        Output a crystal of size n.

        Sample test(s)

        input

        output

        *D*DDD*D*

        input

        output

        **D***DDD*DDDDD*DDD***D**

        input

        output

        ***D*****DDD***DDDDD*DDDDDDD*DDDDD***DDD*****D***


        傳送門:點(diǎn)擊打開(kāi)鏈接

        解題思路:

        水題,推出公式,直接打印即可。

        代碼:

        #include #include 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

        B. Little Pony and Sort by Shift

        time limit per test

        1 second

        memory limit per test

        256 megabytes

        input

        standard input

        output

        standard output

        One 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:

        a1,?a2,?...,?an?→?an,?a1,?a2,?...,?an?-?1.

        Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

        Input

        The first line contains an integer n (2?≤?n?≤?105). The second line contains n integer numbers a1,?a2,?...,?an (1?≤?ai?≤?105).

        Output

        If 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)

        input

        22 1

        output

        input

        31 3 2

        output

        -1

        input

        21 2

        output


        題意:

        給一個(gè)序列,每次可以把序列的最后一個(gè)數(shù)移到最前面,如果可以使序列遞增(嚴(yán)格來(lái)說(shuō)是非遞減),輸入最少移動(dòng)次數(shù),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(kāi)始掃一遍,如果,后面的序列遞增,則輸出其長(zhǎng)度,即為答案,否則,輸出-1。因?yàn)槿绻凑疹}中所述方式移動(dòng),使序列變?yōu)檫f增,序列應(yīng)當(dāng)是本來(lái)就是遞增的,或者是可以分為兩個(gè)連續(xù)的遞增子序列的。

        傳送門:點(diǎn)擊打開(kāi)鏈接

        代碼:

        #include #include #include 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.

        Input

        A single line contains two integers m and n (1?≤?m,?n?≤?105).

        Output

        Output 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)

        input

        6 1

        output

        3.500000000000

        input

        6 3

        output

        4.958333333333

        input

        2 2

        output

        1.750000000000

        傳送門: 點(diǎn)擊打開(kāi)鏈接

        解題思路:

        求數(shù)學(xué)期望,公式P = m - (1/m)^n - (2/m)^n - ... -((m-1)/m)^n

        代碼:

        #include #include #include 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;}

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        文檔

        CodeforcesRound#259(Div.2)題解_html/css

        CodeforcesRound#259(Div.2)題解_html/css_WEB-ITnose:A. Little Pony and Crystal Mine time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (
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        標(biāo)簽: div round 題解
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