3 条题解
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1.高精度算法
//定义和调用省略 inline short reader(){ //快读省略 return it ; } int main(){ //cin >> n //for(i - n) 也可以 for(int i = reader() ; i > 0 ; i --){ for(int j = 0 ; j <= xi ; j ++) x[j] <<= 1 ; //x 乘二 可以使用左移 for(int j = 0 ; j <= xi ; j ++) if(x[j] >= 10) //进位 x[j + 1] += x[j] / 10 , xi += (j == xi) ,//若是最后一位，则数的位数加一 x[j] %= 10 ; for(int j = 0 ; j <= ai ; j ++){ al[j] += x[j] ; //每位加上x if(al[j] >= 10) //进位 al[j + 1] += al[j] / 10 , ai += (j == ai) ,//若是最后一位，则数的位数加一 al[j] %= 10 ; } } int i = ai ; for(; i >= 0 ; i --) putchar(al[i] + '0') ; //putchar输出，其他也可 return 0 ; }第二种，打表！
#include <cstdio> #include <string> using std::string ; string a[101] ; void geter(){ a[1] = "2" , a[2] = "6" , a[3] = "14" , a[4] = "30" , a[5] = "62"; a[6] = "126" , a[7] = "254" , a[8] = "510" , a[9] = "1022" , a[10] = "2046"; a[11] = "4094" , a[12] = "8190" , a[13] = "16382" , a[14] = "32766" , a[15] = "65534"; a[16] = "131070" , a[17] = "262142" , a[18] = "524286" , a[19] = "1048574" , a[20] = "2097150"; a[21] = "4194302" , a[22] = "8388606" , a[23] = "16777214" , a[24] = "33554430" , a[25] = "67108862"; a[26] = "134217726" , a[27] = "268435454" , a[28] = "536870910" , a[29] = "1073741822" , a[30] = "2147483646"; a[31] = "4294967294" , a[32] = "8589934590" , a[33] = "17179869182" , a[34] = "34359738366" , a[35] = "68719476734"; a[36] = "137438953470" , a[37] = "274877906942" , a[38] = "549755813886" , a[39] = "1099511627774" , a[40] = "2199023255550"; a[41] = "4398046511102" , a[42] = "8796093022206" , a[43] = "17592186044414" , a[44] = "35184372088830" , a[45] = "70368744177662"; a[46] = "140737488355326" , a[47] = "281474976710654" , a[48] = "562949953421310" , a[49] = "1125899906842622" , a[50] = "2251799813685246"; a[51] = "4503599627370494" , a[52] = "9007199254740990" , a[53] = "18014398509481982" , a[54] = "36028797018963966" , a[55] = "72057594037927934"; a[56] = "144115188075855870" , a[57] = "288230376151711742" , a[58] = "576460752303423486" , a[59] = "1152921504606846974" , a[60] = "2305843009213693950"; a[61] = "4611686018427387902" , a[62] = "9223372036854775806" , a[63] = "18446744073709551614" , a[64] = "36893488147419103230" , a[65] = "73786976294838206462"; a[66] = "147573952589676412926" , a[67] = "295147905179352825854" , a[68] = "590295810358705651710" , a[69] = "1180591620717411303422" , a[70] = "2361183241434822606846"; a[71] = "4722366482869645213694" , a[72] = "9444732965739290427390" , a[73] = "18889465931478580854782" , a[74] = "37778931862957161709566" , a[75] = "75557863725914323419134"; a[76] = "151115727451828646838270" , a[77] = "302231454903657293676542" , a[78] = "604462909807314587353086" , a[79] = "1208925819614629174706174" , a[80] = "2417851639229258349412350"; a[81] = "4835703278458516698824702" , a[82] = "9671406556917033397649406" , a[83] = "19342813113834066795298814" , a[84] = "38685626227668133590597630" , a[85] = "77371252455336267181195262";; a[86] = "154742504910672534362390526" , a[87] = "309485009821345068724781054" , a[88] = "618970019642690137449562110" , a[89] = "1237940039285380274899124222" , a[90] = "2475880078570760549798248446"; a[91] = "4951760157141521099596496894" , a[92] = "9903520314283042199192993790" , a[93] = "19807040628566084398385987582" , a[94] = "39614081257132168796771975166" , a[95] = "79228162514264337593543950334"; a[96] = "158456325028528675187087900670" , a[97] = "316912650057057350374175801342" , a[98] = "633825300114114700748351602686" , a[99] = "1267650600228229401496703205374" , a[100] = "2535301200456458802993406410750"; } inline short reader(){ } int main(){ geter() ; printf("%s" , a[reader()].c_str()) ; /* //打表代码 freopen("out.txt" , "w" , stdout) ; for(int n = 1 ; n <= 100 ; n ++){ for(int i = 0 ; i < 40 ; i ++) al[i] = x[i] = 0; x[0] = 1 , xi = ai = 0 ; printf("a[%d] = \"" , n) ; for(int i = n ; i > 0 ; i --){ for(int j = 0 ; j <= xi ; j ++) x[j] <<= 1 ; for(int j = 0 ; j <= xi ; j ++) if(x[j] >= 10) x[j + 1] += x[j] / 10 , xi += (j == xi) , x[j] %= 10 ; for(int j = 0 ; j <= ai ; j ++){ al[j] += x[j] ; if(al[j] >= 10) al[j + 1] += al[j] / 10 , ai += (j == ai) , al[j] %= 10 ; } } int i = ai ; for(; i >= 0 ; i --) putchar(al[i] + '0') ; printf("\""); if(n % 5 == 0) printf(";\n ") ; else printf(" , ") ; } */ return 0 ; }-
hetao1156052 @ 2023-7-21 10:30:28
兄弟TLE了啊
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hetao3945559 @ 2024-1-23 23:28:21
我很好奇你那表是怎么算出来又不TLE的。 况且你真有这么大内存吗？
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hetao149375
LV 10
@ 2022-11-5 20:46:42
嗨害嗨，我来了！ 今天咱们讲讲2+2×2+……2×2×2……×2. 看到这题，楼下发话了：“打表不香么？” （真不错） 不不不，这也太low了！ 各位，我向你们讲一个公式： Sn=2^(n+1)-2 (完爆打表者，等比数列求和了解一下） 好，上代码：
a=input() a=int(a) p=2 s=1 for i in range(a+1):#2^（n+1)-2 s *= 2 print(s-2)#首项-末项Loading:12/100……
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hetao149375
@ 2022-11-5 20:51:43
这个方法证明很简单： 设S=2+2×2+……2×2×……×2 则2S=2×2+2×2×2+……2×2×……×2×2 2S-S=S=2+2×2+……2×2×……×2=2^(n+1)-2 即S=2+2×2+……2×2×……×2=2^(n+1)-2 这题采用这个方法，只需要13ms。
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hetao815606
@ 2023-2-19 13:33:03
q=2没用，直接删就行，a = int (input())，要节俭。
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用
python3是不是有点过分了？这里可是c++区！ （虽然我也在用python）
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hetao8951299 LV 10 @ 2023-12-26 19:46:03
#include <bits/stdc++.h> using namespace std; string add(string s,string b) { long long t=0; string res; reverse(s.begin(),s.end()); reverse(b.begin(),b.end()); int len=max(s.size(),b.size()); for(int i=0;i<len;i++) { if(i<s.size()) t+=s[i]-'0'; if(i<b.size()) t+=b[i]-'0'; res+=(char)(t%10+'0'); t/=10; } if(t!=0) res+=(char)(t+'0'); reverse(res.begin(),res.end()); return res; } string mul(string s1,string s2) { string ret; int a[520]={0},b[520]={0},c[1020]={0}; int la=s1.length(),lb=s2.length(); for(int i=0;i<s1.length();i++) { a[s1.length()-i-1]=s1[i]-'0'; } for(int i=0;i<s2.length();i++) { b[s2.length()-i-1]=s2[i]-'0'; } for(int i=0;i<lb;i++) { for(int j=0;j<la;j++) { c[i+j]+=a[j]*b[i]; c[i+j+1]+=c[i+j]/10; c[i+j]%=10; } } int idx=la+lb; while(idx&&!c[idx]) { idx--; } for(int i=idx;i>=0;i--) { ret+=char(c[i]+'0'); } return ret; } string sum="2"; long long n; int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n; for(int i=2;i<=n;i++) { string s="2"; for(int j=1;j<=i-1;j++) { s=mul(s,"2"); } sum=add(sum,s); } cout<<sum; return 0; }
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