bettle_king的博客

二进制枚举: 1代表开 代表不开思路来源于: https://www.bilibili.com/video/BV1wK4y1S7Bd?from=search&seid=8750939626540753681#include<cstdio>#include<iostream>using namespace std;void print(int x){ char a[7][7]={" "," "," "}; for(int i=0;i<7;.

马上要考2021第十二届的蓝桥杯，学校组织了模拟赛，用的是去年的国赛题。做的真的垃圾，于是写一篇总结帖子。目录A: 美丽的2 【水题 / 签到题】B: 扩散 【爆搜】C: 阶乘约数 【数论】E: 玩具蛇 【爆搜】A: 美丽的2 【水题 / 签到题】答案: 563#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorith.

https://www.acwing.com/problem/content/description/2069/#include<iostream>#include<cstring>using namespace std;const int N=40;int n,m;int f[N][N];int main(void){ cin>>n>>m; f[1][1]=1; for(int i=1;i<=n;i++) .

https://www.acwing.com/problem/content/2068/#include<cstdio>#include<iostream>#include<string>#include<cstring>using namespace std;int main(void){ string a; string b=""; cin>>a; for(int i=0;i<a.size();i++) { .

https://www.acwing.com/problem/content/description/2067/#include<iostream>using namespace std;int main(void){ long long a; cin>>a; while(a) { cout<<a<<" "; a=a/2; } return 0;}用右移来解决#include<cstdio>#include&.

http://oj.ecustacm.cn/problem.php?id=1106#include<iostream>#include<cstdio>using namespace std;int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};//上 右 下 左 int main(void){ int n; while(cin>>n) { int s[25][25]={0}; bool m[25][25]={false.

http://oj.ecustacm.cn/problem.php?id=1250#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>using namespace std;int main(void){ int a,b,c; int n; while(cin>>n) { bool flag=false; int max=sqr.

http://oj.ecustacm.cn/problem.php?id=1251注意判断是不是闰年#include<cstdio>#include<iostream>using namespace std;int m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};bool judge(int year){ if(year%400==0||year%4==0&&year%100!=0) return tr.

http://oj.ecustacm.cn/problem.php?id=1252比较花里胡哨的解法:#include<cstdio>#include<iostream>#include<string>#include<sstream>#include<cmath>#include<set>using namespace std;bool judge(string str){ set<char>st;.

http://oj.ecustacm.cn/problem.php?id=1253#include<cstdio>#include<cstring>#include<string>#include<iostream> using namespace std;int ans;int a[15];void dfs(int index){ if(index==13) { int sum=0; for(int i=0;i<13;i.

1254: [蓝桥杯2015初赛]手链样式解析1:用数学思维解决我们先算出总的组合数，接着再按题目检出重复的组合数。有题目不难得出总的组合数为 sum=C123 x C94x C55=27720看题目手链是可以转动的，手链是一个圈。我们先看一个例子:以1为起点得到 123以2为起点得到 231以3为起点得到 312因为手链可以转动，所以这些都是同一种。所以 sum=sum/n=27720/12=2310接下来我们在分析翻转重复的次数我们剩余的sum中有自己对称的图形，还有自己不

http://oj.ecustacm.cn/problem.php?id=1255#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<vector>#include<string>using namespace std;int main(void){ int n; while(cin>>n) {.

http://oj.ecustacm.cn/problem.php?id=1258#include<cstdio>#include<iostream>using namespace std;int ans;int main(void){ for(int i=10000;i<=99999;i++) { bool flag=true; int temp=i; while(temp) { int t=temp%10; if(t==4) f.

http://oj.ecustacm.cn/problem.php?id=1259#include<cstdio>#include<iostream>#include<algorithm>using namespace std;int a[10]={0,1,2,3,4,5,6,7,8,9};int main(void){ /*do { if(a[0]==0) continue; if(a[4]==0) continue; int sum1=.

http://oj.ecustacm.cn/problem.php?id=1260#include<cstdio>#include<iostream>#include<algorithm>using namespace std;int main(void){ /*int sum=2015-1225; for(int i=2;i<=49;i++) { for(int j=i+1;j<=49;j++) { int a=i*(i-2.

http://oj.ecustacm.cn/problem.php?id=1261#include<cstdio>#include<iostream>using namespace std;int w,m,n;int x1,y1;int x2,y2;int main(void){ while(cin>>w>>m>>n) { m--,n--; x1=m/w; x2=n/w; if(x1&a.

http://oj.ecustacm.cn/problem.php?id=1263#include<cstdio>#include<iostream> using namespace std;const int N=1010;int m,n;int main(void){ while(cin>>m>>n) { bool map[N][N]={false}; int s=m+n-1;//横 for(int i=0;i<m;.

http://oj.ecustacm.cn/problem.php?id=1264全排列#include<cstdio>#include<algorithm>using namespace std;int a[]={1,2,3,4,5,6,7,8,9,10};int main(void){ int n=0; do { bool flag=true; for(int i=1;i<=4;i++) { if(a[i]<a[i-1]) .

http://oj.ecustacm.cn/problem.php?id=1265本题用现有的全排列函数计算时间太长，我大致的试了试有很长时间都没有出来答案。只能用dfs + 剪枝来做此题。思路已经有现有的大佬写好了。这里就直接引用了。原文链接自己写的代码:#include<cstdio>#include<string>#include<iostream>#define N 20using namespace std;int a[N][N];.

http://oj.ecustacm.cn/problem.php?id=1266网上找的代码: 自己写了结构体的构造函数，初始化方便。#include<iostream>#include<cstdio>#include<algorithm>#include<string> #include<cstring>#include<queue>using namespace std;int n;int ax[4]=.

http://oj.ecustacm.cn/problem.php?id=12693A+7B+1C =3154A+10B+1C=420A+3B=1053A+9B=315C=2BA=105-3BA+B+C=?105-3B+B+2B=105暴力#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int main(void){ /*int i,j,.

http://oj.ecustacm.cn/problem.php?id=1271这道题最早的思路是把字符串枚举所有的以 8个连续的字符串 并把它们标记为true。根据输入找到就 加1。后来才发现题目给的匹配条件是 两个字符串的只要各个字母的个数相同就是一次匹配。这让我蒙蔽了，这俩字符串该如何匹配啊？ 后来一想只要把每个字符串字典序排序比较就ok了。#include<cstdio>#include<iostream>#include<cstring>.

http://oj.ecustacm.cn/problem.php?id=1274#include<cstdio>#include<algorithm>using namespace std;int main(void){ int i,j,k; int n=0; for(i=1;i<=9;i++) { for(j=0;j<=9;j++) { for(k=0;k<=9;k++) { if(i>j&&.

http://oj.ecustacm.cn/problem.php?id=1275全排列，不过结果要除以10 ，因为5个角旋转，还有镜像(对称)25*2=10#include<cstdio>#include<algorithm>using namespace std;int a[]={1,2,3,4,5,6,8,9,10,12};int main(void){ int n=0; do { int sum1=a[0]+a[1]+a[2]+a[3]; i..

http://oj.ecustacm.cn/problem.php?id=12771+2+3+4+5+6+7+8+9+10=5511+12+13+14+15=6555+65=120>100小明的门牌号一定在这之中。可以猜到它的号码应该靠后，不可能是 1 2 3这些太小的数。多猜测可以猜到 5+6+7+8+9+11+12+13+14+15=100小明的门牌号为10...

http://oj.ecustacm.cn/problem.php?id=1279暴力不多说。#include<cstdio>#include<iostream>using namespace std;int main(void){ int i,j,k; int sum1,sum2; for(i=0;i<=100;i++)//A { for(j=0;j<=100;j++)//B { for(k=0;k<=100;k++)//C .

http://oj.ecustacm.cn/problem.php?id=1280#include<cstdio>#include<iostream>#include<string>#include<sstream>using namespace std;int a[1000];string s;bool judge(long long int sum)//判断是不是回文数 { stringstream s1; s1<<su.

http://oj.ecustacm.cn/problem.php?id=1281#include<cstring>#include<string>#include<cstdio>#include<iostream>using namespace std;int main(void){ string a; int n; cin>>a>>n; for(int i=1;i<=n;i++) { string.

http://oj.ecustacm.cn/problem.php?id=1282#include<cstdio>#include<cstring>#include<string>#include<iostream>using namespace std;int ans;int main(void){ for(int i=1;i<=99;i++) { int j=i+27; int sum=(j%10)*10+j/10; .

http://oj.ecustacm.cn/problem.php?id=1283#include<cstdio>#include<cstring>#include<string>#include<iostream>using namespace std;int main(void){ for(int i=1;i<=100;i++) { int temp=i; int sum=0; while(1) { sum+.

http://oj.ecustacm.cn/problem.php?id=1284#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;int a[10]={0,1,2,3,4,5,6,7,8,9};int main(void){ int n; do { bool flag=true; if(abs(a[0.

http://oj.ecustacm.cn/problem.php?id=1285dfs不是太会写，所以用这个简单但是快的方法了。#include<cstring>#include<cstdio>#include<iostream>#include<string>#include<algorithm>using namespace std;int b[15]={1,2,3,4,5,6,7,8,9,10,11,12,13};int.

http://oj.ecustacm.cn/problem.php?id=1286直接用dfs 是不可以的因为 dfs只可以走L型 不可以走T型那么我们可以先用现有的全排列函数把所有的情况列出来，再判断其可不可以连通。首先关键的一点在于将一位的数组映射到二维。第二点就是如何判断是连通的关于第二点就是用两个数组保存各个的方向。如果可以走的话就把 1 恢复为 0 。如果这个动作按照方向执行了5次说明就是连通的。#include<iostream>#include<cs.

http://oj.ecustacm.cn/problem.php?id=1290#include<cstdio>#include<iostream>using namespace std;int a;long long int sum;int main(void){ for(int i=1;i<=100;i++) { a=(i+1)*i/2; sum+=a; } cout<<sum<<endl; return 0;}.

http://oj.ecustacm.cn/problem.php?id=1291解析: 就是一个全排列。不过有点坑，注意把 后两项合并求，不要分开求。还得判断能不能整除#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int a[9]={1,2,3,4,5,6,7,8,9};int main(void){ int sum1,sum2,sum3; int .

http://oj.ecustacm.cn/problem.php?id=1292#include<cstdio>#include<iostream>using namespace std;int n;int a[100100];int ans;int main(void){ while(cin>>n) { ans=0; for(int i=1;i<=n;i++) cin>>a[i]; for(int i=1;i<.

http://oj.ecustacm.cn/problem.php?id=1293故本题的答案是1728+1125-1原文: https://www.cnblogs.com/-Ackerman/p/12242358.html

http://oj.ecustacm.cn/problem.php?id=1294根据题目不是落入1 就是一个循环圈。如果循环的话说明有的数字出现了两次，那么当某个数字出现两次时就退出循环。并输出最大的值。#include<cstring>#include<string>#include<iostream>#include<cstdio>#include<algorithm>using namespace std;bool h.

http://oj.ecustacm.cn/problem.php?id=1295解析: 题目的意思模糊不清。题目的意思是输入一个N 从 2 ~ N 中 的每一个数 执行冰雹的翻滚，循环直到等于1退出循环。找到这些中冰雹最大的数。#include<cstdio>int main(void){ long long int n; long long int max; long long int temp; while( scanf("%lld",&n) != E.

http://oj.ecustacm.cn/problem.php?id=1296#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<map>using namespace std;int dx[4]={-1,0,1,0};int dy.