Codeforces Round #271 (Div. 2) 解题报告_html/css_WEB-ITnose

题目地址：http://codeforces.com/contest/474

A题：Keyboard

模拟水题。

代码如下：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64char s[]={"qwertyuiopasdfghjkl;zxcvbnm,./"};int main(){    int i, x, j, len;    char c, s1[200];    scanf("%c",&c);    if(c=='L')        x=1;    else        x=-1;        scanf("%s",s1);    len=strlen(s1);    for(i=0;i<len for if printf s break return>  <br> B题：Worms  <p></p>  <p>水题。。</p>  <p>代码如下：</p>  <p></p>  <pre name="code" class="sycode">#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64int dp[1100000];int main(){    int n, m, i, j, sum=0, x;    scanf("%d",&n);    for(i=0;i<n scanf for dp sum while printf return>  <br> C题： Captain Marmot  <p></p>  <p>暴力枚举，共4*4*4*4种情况，对每一种情况分别判断是否是正方形。我居然一直都以为是矩形。。</p>  <p>判断方法：将4条边与两条对角线分别计算出来。然后排序，4个小的肯定是边，2个大的是对角线，然后判断边是否都相等，对角线是否都相等，对角线是否是边的sqrt(2)倍（这里最好是用平方来判断是否是2倍）。然后找出移动次数最少的输出即可。</p>  <p>代码如下：</p>  <p></p>  <pre name="code" class="sycode">#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int mod=1e9+7;struct node{    LL x, y;}t1[5], t2[5], fei[5];node solve(node x, node y, int z){    node t;    t=x;    int i;    for(i=0;i<z x.x="y.y-t.y+y.x;" x.y="t.x-y.x+y.y;" t="x;" return dist a node b ll x="a.x-b.x;" y="a.y-b.y;" judge int i j d sort if main k h min1 scanf while for fei puts else printf>  <br> D题：Flowers  <p></p>  <p>DP，还是水题。。可以这样考虑：</p>  <p>第n个只有两种情况，若第n个是R，那么情况数为dp[n-1]种。若第n个是W，由于W只能连续k个，所以说，第n-k+1至第n个必须都是W，那么此时情况数为dp[n-k]种。所以状态转移方程为：</p>  <p>dp[n]=dp[n-1]+dp[n-k]。</p>  <p>然后用一个数组保存前缀和即可。</p>  <p>代码如下：<br> </p>  <pre name="code" class="sycode">#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int mod=1e9+7;LL dp[110000], sum[110000];int main(){    int i, j, n, k, a, b;    LL x=0;    sum[0]=0;    dp[0]=0;    scanf("%d%d",&n,&k);    for(i=1;i 自己能做出来的只有这么些。。sad。。  <p></p> </algorithm></set></map></queue></ctype.h></math.h></stdlib.h></cstring></string></cstdio></iostream>