紫薯 第三章
3-2 开灯问题
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; using namespace std; int a[5000]; int main() { int n,k,f=1; cin>>n>>k; for(int i=1;i<=n;i++) a[i]=1; for(int i=2;i<=k;i++){ for(int j=i; j<=n; j+=i){ a[j] = !a[j]; } } for(int i=1;i<=n; i++){ if(a[i]){ if(f){ f=0; printf("%d",i); } else printf(" %d", i); } } return 0; } /* 7 3 1 2 3 4 5 6 7 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 0 0 0 1 1 1 */
3-3 🐍形填数
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; int a[N][N]; int n; int main() { cin>>n; int tot = 0; int x, y; tot = a[x=0][y=n-1] = 1; while(tot < n * n){ while(x+1<n && !a[x+1][y]) a[++x][y] = ++tot; while(y-1>=0 && !a[x][y-1]) a[x][--y] = ++tot; while(x-1>=0 && !a[x-1][y]) a[--x][y] = ++tot; while(y+1<n && !a[x][y+1]) a[x][++y] = ++tot; } for(int i=0; i<n; i++){ for(int j=0; j<n; j++){ printf("%3d ",a[i][j]); } cout<<endl; } return 0; }
3-4 竖式问题
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; int a[N][N]; char s[N]; char buf[N]; int n; int main() { scanf("%s",s); int c = 0; for(int abc = 111; abc <= 999; abc++) for(int de = 11; de <= 99; de++){ int x = abc * (de % 10), y = abc * (de / 10), z = abc * de; sprintf(buf, "%d%d%d%d%d", abc, de, x, y, z); int ok = 1; for(int i=0; i<strlen(buf); i++) if(strchr(s, buf[i]) == NULL) ok = 0; // if(ok){ printf("<%d>\n",++c); printf("%5d\nX%4d\n-----\n%5d\n%4d\n-----\n%5d\n\n",abc, de, x, y, z); } } printf("cnt = %d\n",c); return 0; }
3-6: 环状序列
#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; while(n--){ string s, str[150]; cin>>s; int len = s.size(); s = s + s; for(int i=0; i<len; i++){ str[i] = s.substr(i,len); } sort(str,str+len); cout<<str[0]<<endl; } }
例题:3-1 得分
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; int a[N][N]; char s[N]; char buf[N]; int n; int f = 1; int main() { string s; cin>>s; int o = 0, x = 0, sum = 0; for(int i=0; i<s.size(); i++){ if(s[i]=='O') o++; else{ o = 0; } sum += o; } cout<<sum<<endl; return 0; }
3-2:分子量
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; int a[N][N]; char s[N]; char buf[N]; int n; int f = 1; int main() { string s; cin>>s; map<char,double> mp; mp['C'] = 12.01; mp['H'] = 1.008; mp['O'] = 16.00; mp['N'] = 14.01; double sum = 0; for(int i=0; i<s.size(); i++){ if(s[i]=='C'){ if( isdigit(s[i+1]) && !isdigit(s[i+2]) ){ sum += mp[s[i]] * (s[i+1]-'0'); } else if(isdigit(s[i+1]) && isdigit(s[i+2]) ) sum += (mp[s[i]] * ((s[i+1]-'0')*10 + (s[i+2]-'0'))); else if( !isdigit(s[i+1])){ sum += mp[s[i]]; } } if(s[i]=='O'){ if( isdigit(s[i+1]) && !isdigit(s[i+2]) ){ sum += mp[s[i]] * (s[i+1]-'0'); } else if(isdigit(s[i+1]) && isdigit(s[i+2]) ) sum += (mp[s[i]] * ((s[i+1]-'0')*10 + (s[i+2]-'0'))); else if( !isdigit(s[i+1])){ sum += mp[s[i]]; } } if(s[i]=='H'){ if( isdigit(s[i+1]) && !isdigit(s[i+2]) ){ sum += mp[s[i]] * (s[i+1]-'0'); } else if(isdigit(s[i+1]) && isdigit(s[i+2]) ) sum += (mp[s[i]] * ((s[i+1]-'0')*10 + (s[i+2]-'0'))); else if( !isdigit(s[i+1])){ sum += mp[s[i]]; } } if(s[i]=='N'){ if( isdigit(s[i+1]) && !isdigit(s[i+2]) ){ sum += mp[s[i]] * (s[i+1]-'0'); } else if(isdigit(s[i+1]) && isdigit(s[i+2]) ) sum += (mp[s[i]] * ((s[i+1]-'0')*10 + (s[i+2]-'0'))); else if( !isdigit(s[i+1])){ sum += mp[s[i]]; } } } cout<<sum<<endl; return 0; }
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; double a[5] = {12.01, 1.008, 16.00, 14.01}; char s[N]; char buf[5] = {'C','H','O','N'}; int n; int f = 1; double sum = 0; void add(int f, int i) { if(isdigit(s[i+1])) if(isdigit(s[i+2])) sum += a[f] * ( (s[i+1]-'0')*10 + s[i+1]-'0' ); else sum += a[f] * (s[i+1]-'0'); else sum += a[f]; } int main() { int t; scanf("%d",&t); while(t--){ gets(s); sum = 0; for(int i = 0; s[i]; i++){ for(int j = 0; j < 4; j++){ if(s[i] == buf[j]){ add(j, i); } } } printf("%.3f\n",sum); } return 0; }
3-3:数数字
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <cstring> #include <climits> #include <cmath> #include <cctype> const int inf = 0x3f3f3f3f;//1061109567 typedef long long ll; #define N 5000 using namespace std; int a[10005]; char s[N]; int n; int main() { int t; scanf("%d",&t); while(t--){ memset(a,0,sizeof(a)); scanf("%d",&n); for(int i=1; i<=n; i++){ int tmp = i; while(tmp){ a[tmp%10]++; tmp/=10; } } for(int i=0; i<=9; i++){ printf(i == 9 ? "%d\n" : "%d ",a[i]); } } return 0; }
3-4 :周期串
#include<bits/stdc++.h> using namespace std; int main() { int t; cin>>t; while(t--){ string s; cin>>s; int len = s.size(); for(int i=1; i<=len; i++) { if(len % i == 0) { int f = 1; //判断最小周期串是否找到,找到值为1,否则为0 for(int j=i; j<len; j++) { if(s[j] != s[j%i]) { f = 0; break; } } if(f) { printf("%d",i); break; } } } if(t) printf("\n\n"); else printf("\n"); } } /* 从字符串的起始位置遍历，对于字符串长度为N的字符串， 那么周期串的可能长度介于1—N之间，所以可以假设周期串的长度为i， 其中1<=i<=N。满足最小周期串的条件之一是length%i==0， 这是因为如果所述i是字符串的周期串长度， 那么整个字符串可以被划分成整数个长度为i的周期子串， 如果这个整除的条件不成立，说明i也不是字符串的周期串长度。 满足最小周期串的条件之二是， 变量j从i的位置开始一直到字符处结束的位置， 满足(s[j]!=s[j%i])这个条件，这是因为若从0到i-1这个区间组成的子串是该字符串的周期串， 那么从i到length-1的位置是该最小字符串的若干次重复。由于上述i的取值是从1即从最小值开始取值的， 因此当上述两个条件都成立时，即可求得最小周期串长度和最小周期串。 */
