蓝桥杯练习系统： 题1 ： 刚开始的想法是将16进制转为十进制再转为二进制 看了题解之后发现不能这么操作,要先将十六进制转化为二进制之后将2进制转为8进制。 要点string 跟cin的应用。
#include<stdio.h> # include<iostream> #include<string.h> # include<string> using namespace std; int n; int main(){ scanf("%d",&n); while(n -- ){ string st; cin >> st; string a; for(int i = 0;i < st.length();i ++){ if(st[i] == '0') a+= "0000"; else if(st[i] == '1') a+= "0001"; else if(st[i] == '2') a+= "0010"; else if(st[i] == '3') a+= "0011"; else if(st[i] == '4') a+= "0100"; else if(st[i] == '5') a+= "0101"; else if(st[i] == '6') a+= "0110"; else if(st[i] == '7') a+= "0111"; else if(st[i] == '8') a+= "1000"; else if(st[i] == '9') a+= "1001"; else if(st[i] == 'A') a+= "1010"; else if(st[i] == 'B') a+= "1011"; else if(st[i] == 'C') a+= "1100"; else if(st[i] == 'D') a+= "1101"; else if(st[i] == 'E') a+= "1110"; else if(st[i] == 'F') a+= "1111"; } int len = a.length(); if(len %3 == 1){//3位数为一组，凑够一组 a = "00" + a; } else if(len %3 == 2){ a = "0" + a; } int flag = 0; int sum = 0; for(int i = 2;i < a.length();i += 3){ sum = (a[i - 2] - '0')*4+(a[i - 1] - '0')*2+ (a[i] - '0'); if(sum != 0) flag = 1; if(flag == 1){ printf("%d",sum); } } printf("\n"); } return 0; }对于string 和cin 的应用，特地在hdu上找了一题进行练习： 传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1004 要点：
#include<bits/stdc++.h> using namespace std; const int maxn = 1005; map <string,int> num; int n; int main(){ while(scanf("%d",&n)&&(n != 0)){ int cnt = 0; int count[1005]; memset(count,0,sizeof(count)); num.clear(); for(int i = 0;i < n;i ++){ string s; cin >> s; if(num.count(s) == 0){//之前没有在map里面 num[s] = cnt; cnt ++; count[num[s]] ++; } else{ count[num[s]] ++; } } int position = max_element(count,count+cnt) - count; string s; map<string,int>::iterator it = num.begin(); while(it != num.end()){ if(it->second == position){ s = it->first; break; } it ++; } cout << s << endl; } return 0; }map应用扩展：
题2：十六进制转十进制： 知识点三：回文数：
hdu1282 #include<bits/stdc++.h> using namespace std; const int maxn = 1e5; long long n; int a[maxn]; long long ans[maxn]; int top = 0; int ishuiwei(int cnt){ for(int i = 0;i < cnt/2;i ++){ if(a[i] != a[cnt - 1- i]){ return false; } } return true; } int main(){ while(scanf("%lld",&n) != EOF){ memset(ans ,0,sizeof(ans)); top = 0; ans[top ++] = n; int cnt = 0; int cnt2 = 0; while(n != 0){ a[cnt] = n%10; n = n/10; cnt++; } while(ishuiwei(cnt) == 0){ long long sum1 = 0; for(int i = 0;i < cnt;i ++){ sum1 = sum1*10 + a[i]; } long long sum2 = 0; for(int i = cnt - 1;i >= 0;i --){ sum2 = sum2*10 + a[i]; } long long sum = sum1 + sum2; cnt = 0; n = sum; while(n != 0){ a[cnt] = n%10; n = n/10; cnt ++; } ans[top ++] = sum; cnt2 ++; } printf("%d\n",cnt2); printf("%lld",ans[0]); for(int i = 1;i < top;i ++){ printf("--->%lld",ans[i]); } printf("\n"); } return 0; }知识点：大数处理： https://www.cnblogs.com/wzjhoutai/p/7265194.html
hdu 1002 用数组进行模拟 好累 写的很乱 #include<stdio.h> #include<string> #include<string.h> using namespace std; const int maxn = 1005; int t; char s1[maxn]; char s2[maxn]; int a1[maxn]; int a2[maxn]; int ans[maxn]; int main(){ scanf("%d",&t); int cnttt = 1; while(t -- ){ scanf("%s",&s1); scanf("%s",&s2); int len1 = strlen(s1); int len2 = strlen(s2); int b = 0; if(cnttt > 1) printf("\n"); printf("Case %d:\n",cnttt); cnttt ++; for(int i = 0;i < len1;i ++){ a1[i] = s1[i] - '0'; printf("%d",a1[i]); } printf(" + "); for(int i = 0;i < len2;i ++){ a2[i] = s2[i] - '0'; printf("%d",a2[i]); } printf(" = "); int ma = 0,mi = 0; if(len1 > len2){ b = 1; ma = len1; mi = len2; } else if(len1 < len2){ b = 2; ma = len2; mi = len1; } else{ b = 0; ma = len2; mi = len1; } // printf("b = %d\n",b); int cinn = 0;//表示进位 for(int i = 0;i < mi;i ++){ int a = a1[len1 - 1 - i] +a2[len2 - 1 - i] + cinn; cinn = 0; if(a >= 10){ cinn = 1; a = a - 10; } ans[ma - 1 - i] = a; // printf("ans = %d\n",ans[ma - 1 - i]); } if(b == 1){ for(int i = len1 - mi - 1;i >= 0;i --){ int a = a1[i] + cinn; cinn = 0; if(a >= 10){ cinn = 1; a = a - 10; } ans[i] = a; } } else if(b == 2){ for(int i = len2 - mi - 1;i >= 0;i --){ int a = a2[i] + cinn; // printf(" i = %d,b = 2:a = %d\n",i,a); cinn = 0; if(a >= 10){ cinn = 1; a = a - 10; } ans[i] = a; } } else if(b == 0){ if(cinn == 1) printf("1"); int flag = 0;//排掉不为0 for(int i = 0;i < ma;i ++){ if(ans[i] != 0){ flag = 1; } if(flag == 1) printf("%d",ans[i]); } printf("\n"); continue; } int flag = 0; if(cinn == 1){ printf("1"); flag = 1; } for(int i = 0;i < ma;i ++){ if(ans[i] != 0){ flag = 1; } if(flag == 1) printf("%d",ans[i]); } printf("\n"); } return 0; } /* 1 119999 999 */ hdu 1042 传送门：https://www.cnblogs.com/blumia/p/hdu1042.html 我感觉这个博主的模板写的挺好的 #include<stdio.h> using namespace std; const int maxn = 10005; int n; char s[maxn]; void factorial(){ int a[500000] = {1}; int carry,j; int digit = 1; for(int i = 2;i <= n;i ++){ for(j = 0,carry = 0;j < digit;j ++){ int temp = a[j] * i + carry;//模拟相乘+进位 a[j] = temp % 10 ; carry = temp / 10; } //相乘结束，开始处理对应的进位 while(carry != 0){ a[++digit - 1] = carry%10; carry = carry/ 10; } } //开始打印； for(int i = digit - 1;i >= 0;i --){ printf("%d",a[i]); } printf("\n"); } int main(){ while(scanf("%d",&n) != EOF){ factorial(); } return 0; } hdu 1047 这个方法比我自己写的好很多，还是应该多看别人的实例。 #include<stdio.h> #include<cstring> #include<iostream> using namespace std; const int maxn =5; int t ; string myadd(string str1,string str2){ int len1 = str1.length(); int len2 = str2.length(); string ans; // printf("len1 = %d,len2 = %d\n",len1,len2); if(len1 < len2){//字符串1的长度小于字符串2的长度，将其 补0 for(int i = 0;i < len2 - len1;i ++){ str1 = '0' + str1; } } else{ for(int i = 0;i < len1 - len2;i ++){ str2 = '0' + str2; } } // cout << "str1:"<<str1 << endl; // cout << "str2:"<<str2 << endl; int carry = 0; len1 = str1.length(); for(int i = len1 - 1;i >= 0;i --){//开始诸位相加 int temp = (str1[i] - '0' ) + (str2[i] - '0') + carry; // printf("temp = %d\n",temp); carry = temp / 10; temp = temp %10; ans = char(temp + '0') + ans; } if(carry != 0){ ans = char(carry + '0') + ans; } // cout << "ans:"<<ans << endl; return ans; } int main(){ scanf("%d",&t); while(t -- ){ string sum = "0"; string s; while(cin >> s){ if(s == "0") break; sum = myadd(sum,s); // cout << sum << endl; } cout << sum << endl; if(t > 0) cout << endl; } } /* 2 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0 123456789012345678901234567890000000000 123456789012345678901234567890000000000 123456789012345678901234567890000000000 0 */ hdu1715 试图打表，然后失败 tle代码： #include<stdio.h> #include<string> #include<string.h> #include<cstring> #include<iostream> using namespace std; const int maxn = 1005; long long dp[maxn]; int t; string myadd(string str1,string str2){ string str; int len1 = str1.length(); int len2 = str2.length(); if(len1 > len2){ for(int i = 0 ;i < len1 - len2;i ++){ str2 = '0'+ str2; } } else{ for(int i = 0 ;i < len2 - len1;i ++){ str1 = '0'+ str1; } } len1 = str1.length(); // printf("len1 = %d,len2 = %d",len1,len2); int carry = 0; for(int i = len1 - 1;i >= 0;i -- ){ int temp = (str1[i] - '0') + (str2[i] - '0') + carry; carry = temp/10; temp = temp%10; str = char(temp + '0')+ str; } if(carry != 0){ str = char(carry + '0')+ str; } // cout << "str:" << str << endl; return str; } int main(){ //得到n< 80之下的答案 dp[79] = 23416728348467685 dp[0] = 1; dp[1] = 1; for(int i = 2;i < 80;i ++){ dp[i] = dp[i - 1]+ dp[i - 2]; } string str1 = "14472334024676221";//f[79]的数 string str2 = "23416728348467685";//f[80]的数 scanf("%d",&t); int n; while(t --){ scanf("%d",&n); if(n <= 80){ printf("%lld\n",dp[n-1]); continue; } else{ int t = n - 80; // printf("t = %d\n",t); string sum; while(t --){ sum = myadd(str1,str2); // cout <<"sum1:" << sum << endl; str1 = str2; str2 = sum; } cout << sum << endl; } } return 0; } /* 13 81 14 55 234 999 1 2 3 4 5 6 7 8 */最后还是打表过的，窒息：
#include<stdio.h> #include<string> #include<string.h> #include<cstring> #include<iostream> using namespace std; const int maxn = 1005; long long dp[maxn]; int t; string ans[maxn]; string myadd(string str1,string str2){ string str; int len1 = str1.length(); int len2 = str2.length(); if(len1 > len2){ for(int i = 0 ;i < len1 - len2;i ++){ str2 = '0'+ str2; } } else{ for(int i = 0 ;i < len2 - len1;i ++){ str1 = '0'+ str1; } } len1 = str1.length(); // printf("len1 = %d,len2 = %d",len1,len2); int carry = 0; for(int i = len1 - 1;i >= 0;i -- ){ int temp = (str1[i] - '0') + (str2[i] - '0') + carry; carry = temp/10; temp = temp%10; str = char(temp + '0')+ str; } if(carry != 0){ str = char(carry + '0')+ str; } // cout << "str:" << str << endl; return str; } int main(){ //得到n< 80之下的答案 dp[79] = 23416728348467685 ans[1] = "1"; ans[2] = "1"; string str1 ;//f[79]的数 string str2;//f[80]的数 for(int i = 3;i <= 1000;i ++){ str1 = ans[i - 2]; str2 = ans[i - 1]; ans[i] = myadd(str1,str2); } scanf("%d",&t); int n; while(t --){ scanf("%d",&n); cout << ans[n] << endl; } return 0; } /* 13 81 14 55 234 999 1 2 3 4 5 6 7 8 */ hdu 1063 快速幂。告辞了，不会写，等裤子帮我写吧 真香，然后7.2一边玩一边写 我感觉写的很对啊，但是就是跑到一半程序就崩溃了 #include<stdio.h> #include<string> #include<string.h> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int maxn = 605; char s[40]; int n; int ans[maxn]; string str1; string str2; string sum; //乘法模板 string myplus(string str1,string str2){ sum.clear(); memset(ans,0,sizeof(ans)); int carry;//表示进位 for(int i = 0;i < str1.length();i ++){ for(int j = 0;j < str2.length() ;j ++){ ans[i + j] += (str1[i] - '0')*(str2[j] - '0'); // printf("a [%d] = %d * %d",i + j,(str1[i] - '0'),(str2[j] - '0')); // printf("= %d\n",ans[i + j]); } } for(int i = 0;i < str1.length() + str2.length() - 1;i ++){ carry = ans[i] / 10; ans[i + 1] += carry; ans[i] %= 10; sum = char(ans[i] + '0') + sum; } while(carry != 0){ sum = char((carry % 10)+ '0') + sum; carry = carry/10; } return sum; } int main(){ while(scanf("%s %d",&s,&n) != EOF){ str1.clear(); str2.clear(); int len = strlen(s); int xs = 0; int isxs = 0; int digit = str1.length(); for(int i = 0;i < len;i ++){//统计小数的个数 if(s[i] == '.'){ xs = i; break; } } int flag = 0; for(int i = len - 1;i >= 0;i --){//统计小数的个数 if(s[i] != '0'){ flag = 1; } if(flag == 1) if((s[i] >= '0')&&(s[i] <= '9')){ str1 += s[i]; } } str2 = str1; // printf("len = %d,strlen = %d %d\n",len,str1.length(),str2.length()); //得到小数的位数 xs = str1.length() - xs ; // printf("len - xs - 1 = %d\n",len - xs - 1); if(len - xs - 1 == 1){ if(s[0] == '0') isxs = 1; } string sum; n = n -1; int point = xs; // printf("point = %d\n",point); while(n -- ){ // cout << "str1 : " << str1 << endl; // cout << "str2 : " << str2 << endl; sum = myplus(str2,str1); reverse(sum.begin(),sum.end()); str2 = sum; point += xs; } reverse(sum.begin(),sum.end()); // printf("isxs = %d\n",isxs); if(isxs == 0) for(int i = 0;i < sum.length()- point;i ++){ cout << sum[i]; } cout << "." ; for(int i = sum.length()- point;i < sum.length();i ++){ cout << sum[i]; } cout << endl ; } return 0; } /* 2.2 3 */后来爸爸靠自己AC了，tmd
#include<stdio.h> #include<string> #include<string.h> #include<cstring> #include<iostream> #include<algorithm> #include<math.h> using namespace std; const int maxn = 605; char s[40]; int n; int ans[maxn]; string str1; string str2; string sum; //乘法模板 string myplus(string str1,string str2){ sum.clear(); memset(ans,0,sizeof(ans)); int carry;//表示进位 for(int i = 0;i < str1.length();i ++){ for(int j = 0;j < str2.length() ;j ++){ ans[i + j] += (str1[i] - '0')*(str2[j] - '0'); // printf("a [%d] = %d * %d",i + j,(str1[i] - '0'),(str2[j] - '0')); // printf("= %d\n",ans[i + j]); } } for(int i = 0;i < str1.length() + str2.length() - 1;i ++){ carry = ans[i] / 10; ans[i + 1] += carry; ans[i] %= 10; sum = char(ans[i] + '0') + sum; } while(carry != 0){ sum = char((carry % 10)+ '0') + sum; carry = carry/10; } // cout << " sum:" << sum << endl; return sum; } int main(){ while(scanf("%s %d",&s,&n) != EOF){ int iszs = 0; int isxs = 0; if(n == 0){ cout << "1" << endl; continue; } int len = strlen(s); str1.clear(); //筛掉前面的0 int flag = 0; for(int i = 0;i < len;i ++){ if(s[i] != '0'){ flag = 1; } if(flag == 1){ str1 += s[i]; } } str2.clear(); str2 = str1; //统计小数位数 int pos = -1; for(int i = 0;i < str1.length();i ++){ if(str1[i] == '.'){ pos = i; break; } } if(pos == -1){ iszs = 1; } if(iszs == 0){ // cout << str1 << endl; //筛掉后面的0 flag = 0; str2.clear(); for(int i = str1.length() - 1;i >= 0;i --){ if(str1[i] != '0'){ // cout << str1[i] << endl; flag = 1; } if(flag == 1){ str2 = str1[i] + str2; } } str1 = str2; } //统计小数位数 pos = -1; for(int i = 0;i < str1.length();i ++){ if(str1[i] == '.'){ pos = i; break; } } // printf("pos = %d\n",pos); //没有发现小数点，是整数 if((pos == -1)||(pos == str1.length() - 1)){ iszs = 1; } //第一位就是小数点， 是小数 if(pos == 0){ isxs = 1; } int xsnum = 0; // printf("strlen = %d,pos = %d\n",str1.length(),pos); xsnum = str1.length() -1 - pos ; //筛掉小数点 str2.clear(); for(int i = str1.length() - 1;i >= 0;i --){ if((str1[i] >= '0')&&(str1[i] <= '9')){ str2 += str1[i]; } } str1 = str2; str2 = "1"; // cout << "srt1:" << str1 << endl; string sum; int point = 0; // printf("point = %d,xsnum = %d\n",point,xsnum); while(n -- ){ // cout << "str1 : " << str1 << endl; // cout << "str2 : " << str2 << endl; sum = myplus(str2,str1); reverse(sum.begin(),sum.end()); str2 = sum; point += xsnum; } reverse(sum.begin(),sum.end()); // cout << "sum:" << sum << endl; if(iszs == 1){ cout << sum << endl; continue; } if(isxs == 0){ for(int i = 0;i < sum.length()- point;i ++){ cout << sum[i]; } cout << "." ; for(int i = sum.length()- point;i < sum.length();i ++){ cout << sum[i]; } cout << endl; continue; } if(isxs == 1){ int t = sum.length(); // printf("point = %d,sum.length() = %d\n",point,sum.length()); if(point > t){ for(int i = 0;i < point - t;i ++){ // cout << "sum:" << sum << endl; sum = '0' + sum; } } cout << "." << sum << endl; } } return 0; } /* 001.10 2 000.01 0 011.01 3 000.00 2 000000 2 00000. 2 0.0100 2 99.999 0 */ hdu1316 大数计算 斐波那契数 #include<stdio.h> #include<bits/stdc++.h> using namespace std; const int maxn = 105; const int maxm = 500; char a[maxn],b[maxn]; string fb[maxm]; string countfb(string str1,string str2){ string sum; int len1 = str1.length(); int len2 = str2.length(); if(len1 > len2) for(int i = 0;i < len1 - len2;i ++) str2 = '0' + str2; else for(int i = 0;i < len2 - len1;i ++) str1 = '0' + str1; int carry = 0; for(int i = str1.length() - 1;i >= 0;i -- ){ int temp = (str1[i] -'0') + (str2[i] -'0') + carry; sum = char((temp % 10) + '0')+sum; carry = temp / 10; } if(carry != 0){ sum = char(carry + '0') +sum; } return sum; } int smallthanA(){ int cnt = 0; int lena = strlen(a); int flag = 0; int cnt2 = 0; for(int i = 1;i < maxm;i ++){ flag = 0; if(lena > fb[i].length()){ // cout <<"fb1:"<<fb[i]<< endl; cnt ++; continue; } else if(lena < fb[i].length()) break; else{ if(a == fb[i]) continue; for(int j = 0;j < lena;j ++){ if((a[j]- '0') > (fb[i][j] - '0')) break; if((a[j]- '0') < (fb[i][j] - '0')){ flag = 1; break; } } if(flag == 0){ // cout <<"fb:2"<<fb[i]<< endl; cnt++; } if(flag == 1){ break; } } } return cnt; } int largethanB(){ int cnt = 0; int lenb = strlen(b); int flag = 0; for(int i = maxm - 1;i >= 0;i --){ flag = 0; if(lenb < fb[i].length()){ cnt ++; continue; } else if(lenb > fb[i].length()) break; else{ if(b == fb[i]) continue; for(int j = 0;j < lenb;j ++){ if((b[j]- '0') < (fb[i][j] - '0')) break; if((b[j]- '0') > (fb[i][j] - '0')){ flag = 1; break; } } if(flag == 0){ cnt++; } if(flag == 1){ break; } } } return cnt; } int main(){ fb[1] = "1"; fb[2] = "2"; for(int i = 3;i < maxm;i ++){ fb[i] = countfb(fb[i - 1],fb[i - 2]); // cout << fb[i] << endl; } while(scanf("%s %s",&a,&b)){ if((a[0] == '0')&&(b[0] == '0')){ break; } int cnt1 = smallthanA(); // printf("cnt = %d\n",cnt1); int cnt2 = largethanB(); // printf("cnt = %d\n",cnt2); int ans = maxm - 1 - cnt1 - cnt2; printf("%d\n",ans); } return 0; } /* 0 1 */ hdu 1753https://paste.ubuntu.com/p/5xgtzjZgCj/
