标签：alt mit define sizeof asterisk 地方 one mis class
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 766 Accepted Submission(s):
190
29697684282993
transcendental
28011622636823854456520
computationally
zzzzzzzzzzzzzzzzzzzz
*
elementary 29,697,684,282,993
transcendental 51,346,529,199,396,181,750
prestidigitation 28,011,622,636,823,854,456,520
computationally 232,049,592,627,851,629,097
zzzzzzzzzzzzzzzzzzzz 20,725,274,851,017,785,518,433,805,270
（注意：数字是从第23列开始写，之前空格不能丢了）
1 # include <stdio.h> 2 # include <string.h> 3 # include <stdlib.h> 4 # include <iostream> 5 # include <fstream> 6 # include <vector> 7 # include <queue> 8 # include <stack> 9 # include <map> 10 # include <math.h> 11 # include <algorithm> 12 using namespace std; 13 # define pi acos(-1.0) 14 # define mem(a,b) memset(a,b,sizeof(a)) 15 # define FOR(i,a,n) for(int i=a; i<=n; ++i) 16 # define For(i,n,a) for(int i=n; i>=a; --i) 17 # define FO(i,a,n) for(int i=a; i<n; ++i) 18 # define Fo(i,n,a) for(int i=n; i>a ;--i) 19 typedef long long LL; 20 typedef unsigned long long ULL; 21 22 //string比较函数:相等返回0，str1>str2返回1，str1<str2返回-1. 23 int Compare(string str1,string str2) 24 { 25 if(str1.length() > str2.length()) return 1; 26 else if(str1.length() < str2.length()) return -1; 27 else return str1.compare(str2); 28 } 29 30 string Big_Plus(string str1,string str2) 31 { 32 string ans; 33 int len1=str1.length(); 34 int len2=str2.length(); 35 //将长度较小的前面补0，使两个string长度相同 36 if(len1<len2){ 37 for(int i=1;i<=len2-len1;i++){ 38 str1="0"+str1; 39 } 40 }else { 41 for(int i=1;i<=len1-len2;i++){ 42 str2="0"+str2; 43 } 44 } 45 int len=max(len1,len2); 46 int carry=0; 47 for(int i=len-1;i>=0;i--){ 48 int tmp=str1[i]-‘0‘+str2[i]-‘0‘+carry; 49 carry=tmp/10; 50 tmp%=10; 51 ans=char(tmp+‘0‘)+ans; 52 } 53 if(carry) ans=char(carry+‘0‘)+ans; 54 return ans; 55 } 56 57 //支持大数减小数 58 string Big_Sub(string str1,string str2) 59 { 60 string ans; 61 int carry=0; 62 int difference=str1.length()-str2.length();//长度差 63 for(int i=str2.length()-1;i>=0;i--){ 64 if(str1[difference+i]<str2[i]+carry){ 65 ans=char(str1[difference+i]+10-str2[i]-carry+‘0‘)+ans; 66 carry=1; 67 }else { 68 ans=char(str1[difference+i]-str2[i]-carry+‘0‘)+ans; 69 carry=0; 70 } 71 } 72 for(int i=difference-1;i>=0;i--){ 73 if(str1[i]-carry>=‘0‘){ 74 ans=char(str1[i]-carry)+ans; 75 carry=0; 76 }else { 77 ans=char(str1[i]-carry+10)+ans; 78 carry=1; 79 } 80 } 81 //去除前导0 82 ans.erase(0,ans.find_first_not_of(‘0‘)); 83 if(ans.empty()) ans="0"; 84 return ans; 85 } 86 87 string Big_Mul(string str1,string str2) 88 { 89 string ans; 90 int len1=str1.length(); 91 int len2=str2.length(); 92 for(int i=len2-1;i>=0;i--){ 93 string tmpstr=""; 94 int data=str2[i]-‘0‘; 95 int carry=0; 96 if(data!=0){ 97 for(int j=1;j<=len2-1-i;j++){ 98 tmpstr+="0"; 99 } 100 for(int j=len1-1;j>=0;j--){ 101 int t=data*(str1[j]-‘0‘)+carry; 102 carry=t/10; 103 t%=10; 104 tmpstr=char(t+‘0‘)+tmpstr; 105 } 106 if(carry!=0) tmpstr=char(carry+‘0‘)+tmpstr; 107 } 108 ans=Big_Plus(ans,tmpstr); 109 } 110 ans.erase(0,ans.find_first_not_of(‘0‘)); 111 if(ans.empty()) ans="0"; 112 return ans; 113 } 114 115 //正数相除，商为quotient,余数为residue 116 117 void Big_Div(string str1,string str2,string& quotient,string& residue) 118 { 119 quotient=residue="";//商和余数清空 120 if(str2=="0"){//；判断除数是否为0 121 quotient=residue="ERROR"; 122 return; 123 } 124 if(str1=="0"){//判断被除数是否为0 125 quotient=residue="0"; 126 return; 127 } 128 int res=Compare(str1,str2); 129 if(res<0){//被除数小于除数 130 quotient="0"; 131 residue=str1; 132 return; 133 }else if(res==0){ 134 quotient="1"; 135 residue="0"; 136 return ; 137 }else { 138 int len1=str1.length(); 139 int len2=str2.length(); 140 string tmpstr; 141 tmpstr.append(str1,0,len2-1);//将str1的前len2位赋给tmpstr 142 for(int i=len2-1;i<len1;i++){ 143 tmpstr=tmpstr+str1[i];//被除数新补充一位 144 tmpstr.erase(0,tmpstr.find_first_not_of(‘0‘));//去除前导0 145 if(tmpstr.empty()) tmpstr="0"; 146 for(char ch=‘9‘;ch>=‘0‘;ch--){//试商 147 string tmp,ans; 148 tmp=tmp+ch; 149 ans=Big_Mul(str2,tmp);//计算乘积 150 if(Compare(ans,tmpstr)<=0){//试商成功 151 quotient=quotient+ch; 152 tmpstr=Big_Sub(tmpstr,ans);//减掉乘积 153 break; 154 } 155 } 156 } 157 residue=tmpstr; 158 } 159 quotient.erase(0,quotient.find_first_not_of(‘0‘)); 160 if(quotient.empty()) quotient="0"; 161 } 162 163 string change(int num) 164 { 165 string n=""; 166 stack<char>M; 167 while(num>0) 168 { 169 M.push(num%10+‘0‘); 170 num/=10; 171 } 172 while(!M.empty()) 173 { 174 n+=M.top(); 175 M.pop(); 176 } 177 return n; 178 } 179 180 int change(string num) 181 { 182 int n=num[0]-‘0‘; 183 for(int i=1;i<num.size();i++) 184 n=n*10+num[i]-‘0‘; 185 return n; 186 } 187 188 int main() 189 { 190 //freopen("in.txt", "r", stdin); 191 string s; 192 while(cin>>s,s[0]!=‘*‘) 193 { 194 if(‘a‘<=s[0]&&s[0]<=‘z‘) 195 { 196 string num=change(s[0]-‘a‘+1); 197 for(int i=1;i<s.size();i++) 198 { 199 num=Big_Plus(Big_Mul(num,change(26)),change(s[i]-‘a‘+1)); 200 } 201 cout<<s; 202 for(int i=s.size()+1;i<23;i++)cout<<‘ ‘; 203 for(int i=0;i<num.size();i++) 204 { 205 if(i) 206 printf((num.size()-i)%3==0?",%c":"%c",num[i]); 207 else 208 printf("%c",num[i]); 209 } 210 cout<<endl; 211 } 212 else if(‘0‘<=s[0]&&s[0]<=‘9‘) 213 { 214 stack<char>M; 215 string str1,str2;//商和余数 216 string str="",str3=s; 217 while(Compare(str3,"0")>0)//str1>0 218 { 219 Big_Div(str3,"26",str1,str2); 220 str3=str1; 221 if(str2=="0") 222 { 223 M.push(‘z‘); 224 str3=Big_Sub(str3,"1"); 225 } 226 else 227 { 228 M.push(change(str2)+‘a‘-1); 229 } 230 } 231 while(!M.empty()) 232 { 233 str+=M.top(); 234 M.pop(); 235 } 236 cout<<str; 237 for(int i=str.size()+1;i<23;i++)cout<<‘ ‘; 238 for(int i=0;i<s.size();i++) 239 { 240 if(i) 241 printf((s.size()-i)%3==0?",%c":"%c",s[i]); 242 else 243 printf("%c",s[i]); 244 } 245 cout<<endl; 246 } 247 } 248 return 0; 249 }
HDU 1314 Numerically Speaking（大数加减乘除+另类二十六进制互相转换）
标签：alt mit define sizeof asterisk 地方 one mis class
原文地址：http://www.cnblogs.com/teble/p/7224597.html
