Sir_Kay

Kaysman Official Website


  • 首页

  • 分类

  • 标签

  • 归档

  • 搜索

Ural 1248 Sequence Sum 题解

发表于 2019-08-18 更新于 2019-08-31 分类于 算法 , Ural
本文字数: 7k 阅读时长 ≈ 6 分钟

Ural 1248 Sequence Sum 题解

题意

给定$n$个用科学计数法表示的实数$(10^{-100}\sim10^{100})$,输出它们的和。

Tip: 一个实数可以用科学计数法表示为$x\times10^y$,其中$1\le x<10$ $x$为实数,$y$是整数。输入时表示为$xey$。保证输入的实数有$19$位有效数字。输出时用科学计数法,必须包括$19$位正确数字。

题解

  1. 读入字符串,把它们换成正常形态放入数组。
  2. 小数点对齐。
  3. 计算。
  4. 换成科学计数法输出。

Tip: 口胡得很简单,写起来可能有点麻烦,要仔细。

程序

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
// #pragma GCC optimize(2)
// #pragma G++ optimize(2)
// #pragma comment(linker,"/STACK:102400000,102400000")

// #include <bits/stdc++.h>
#include <map>
#include <set>
#include <list>
#include <array>
#include <cfenv>
#include <cmath>
#include <ctime>
#include <deque>
#include <mutex>
#include <queue>
#include <ratio>
#include <regex>
#include <stack>
#include <tuple>
#include <atomic>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <chrono>
#include <cstdio>
#include <cwchar>
#include <future>
#include <limits>
#include <locale>
#include <memory>
#include <random>
#include <string>
#include <thread>
#include <vector>
#include <cassert>
#include <climits>
#include <clocale>
#include <complex>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdint>
#include <cstdlib>
#include <cstring>
#include <ctgmath>
#include <cwctype>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <ccomplex>
#include <cstdbool>
#include <iostream>
#include <typeinfo>
#include <valarray>
#include <algorithm>
#include <cinttypes>
#include <cstdalign>
#include <stdexcept>
#include <typeindex>
#include <functional>
#include <forward_list>
#include <system_error>
#include <unordered_map>
#include <unordered_set>
#include <scoped_allocator>
#include <condition_variable>
// #include <conio.h>
// #include <windows.h>
using namespace std;

typedef long long LL;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef float fl;
typedef double ld;
typedef long double LD;
typedef pair<int,int> pii;
#if (WIN32) || (WIN64) || (__WIN32) || (__WIN64) || (_WIN32) || (_WIN64) || (WINDOWS)
#define lld "%I64d"
#define llu "%I64u"
#else
#define lld "%lld"
#define llu "%llu"
#endif
#define ui(n) ((unsigned int)(n))
#define LL(n) ((long long)(n))
#define ull(n) ((unsigned long long)(n))
#define fl(n) ((float)(n))
#define ld(n) ((double)(n))
#define LD(n) ((long double)(n))
#define char(n) ((char)(n))
#define Bool(n) ((bool)(n))
#define fixpoint(n) fixed<<setprecision(n)

const int INF=1061109567;
const int NINF=-1044266559;
const LL LINF=4557430888798830399;
const ld eps=1e-15;
#define MOD (1000000007)
#define PI (3.1415926535897932384626433832795028841971)

/*
#define MB_LEN_MAX 5
#define SHRT_MIN (-32768)
#define SHRT_MAX 32767
#define USHRT_MAX 0xffffU
#define INT_MIN (-2147483647 - 1)
#define INT_MAX 2147483647
#define UINT_MAX 0xffffffffU
#define LONG_MIN (-2147483647L - 1)
#define LONG_MAX 2147483647L
#define ULONG_MAX 0xffffffffUL
#define LLONG_MAX 9223372036854775807ll
#define LLONG_MIN (-9223372036854775807ll - 1)
#define ULLONG_MAX 0xffffffffffffffffull
*/

#define MP make_pair
#define MT make_tuple
#define All(a) (a).begin(),(a).end()
#define pall(a) (a).rbegin(),(a).rend()
#define log2(x) log(x)/log(2)
#define Log(x,y) log(x)/log(y)
#define SZ(a) ((int)(a).size())
#define rep(i,n) for(int i=0;i<((int)(n));i++)
#define rep1(i,n) for(int i=1;i<=((int)(n));i++)
#define repa(i,a,n) for(int i=((int)(a));i<((int)(n));i++)
#define repa1(i,a,n) for(int i=((int)(a));i<=((int)(n));i++)
#define repd(i,n) for(int i=((int)(n))-1;i>=0;i--)
#define repd1(i,n) for(int i=((int)(n));i>=1;i--)
#define repda(i,n,a) for(int i=((int)(n));i>((int)(a));i--)
#define repda1(i,n,a) for(int i=((int)(n));i>=((int)(a));i--)
#define FOR(i,a,n,step) for(int i=((int)(a));i<((int)(n));i+=((int)(step)))
#define repv(itr,v) for(__typeof((v).begin()) itr=(v).begin();itr!=(v).end();itr++)
#define repV(i,v) for(auto i:v)
#define repE(i,v) for(auto &i:v)
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x) MS(x,0)
#define MINF(x) MS(x,63)
#define MCP(x,y) memcpy(x,y,sizeof(y))
#define sqr(x) ((x)*(x))
#define UN(v) sort(All(v)),v.erase(unique(All(v)),v.end())
#define filein(x) freopen(x,"r",stdin)
#define fileout(x) freopen(x,"w",stdout)
#define fileio(x)\
freopen(x".in","r",stdin);\
freopen(x".out","w",stdout)
#define filein2(filename,name) ifstream name(filename,ios::in)
#define fileout2(filename,name) ofstream name(filename,ios::out)
#define file(filename,name) fstream name(filename,ios::in|ios::out)
#define Pause system("pause")
#define Cls system("cls")
#define fs first
#define sc second
#define PC(x) putchar(x)
#define GC(x) x=getchar()
#define Endl PC('\n')
#define SF scanf
#define PF printf

inline int Read()
{
int X=0,w=0;char ch=0;while(!isdigit(ch)){w|=ch=='-';ch=getchar();}while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
return w?-X:X;
}
inline void Write(int x){if(x<0)putchar('-'),x=-x;if(x>9)Write(x/10);putchar(x%10+'0');}

inline LL powmod(LL a,LL b){LL RES=1;a%=MOD;assert(b>=0);for(;b;b>>=1){if(b&1)RES=RES*a%MOD;a=a*a%MOD;}return RES%MOD;}
inline LL gcdll(LL a,LL b){return b?gcdll(b,a%b):a;}
const int dx[]={0,1,0,-1,1,-1,-1,1};
const int dy[]={1,0,-1,0,-1,-1,1,1};
/************************************************************Begin************************************************************/
const int maxn=1010;
const int Dot=500;

int n,dig[maxn][maxn];

inline int Atoi(string x)
{
int res=0;
repV(i,x) if(isdigit(i)) res=(res+i-'0')*10;res/=10;
return (x[0]=='-'?-res:res);
}

inline void func(int k,string x)
{
int e=x.find("e");

string a=x.substr(0,e),b=x.substr(e+1);
if(a.find(".")==string::npos) a+='.';

rep(i,101) a='0'+a;
rep(i,101) a+='0';

int dot=a.find(".")+Atoi(b)-(b[0]=='-'?1:0);

int hav=0;
repd(i,dot+1)
{
if(a[i]=='.')
{
hav=1;
continue;
}

dig[k][Dot-(dot-i+1)+hav]=a[i]-'0';
}

hav=0;
repa(i,dot+1,a.size())
{
if(a[i]=='.')
{
hav=1;
continue;
}

dig[k][Dot+(i-dot)-hav]=a[i]-'0';
}
}

int main()
{
cin>>n;
rep(i,n)
{
string x;cin>>x;
func(i,x);
}

rep(i,n) rep(j,1000) dig[n][j]+=dig[i][j];

repd(j,1000)
{
while(dig[n][j]>=10)
{
int hav=0;
if(j-1==Dot) hav=1;

dig[n][j-1-hav]+=dig[n][j]/10;
dig[n][j]%=10;
}
}

int w=INF,e=0;
rep(i,Dot) if(dig[n][i])
{
w=i;
e=Dot-w-1;
break;
}

if(w==INF)
{
repa(i,Dot+1,1000) if(dig[n][i])
{
w=i;
e=i-Dot;
break;
}
e=-e;
}

cout<<dig[n][w]<<'.';
repa(i,w+1,w+20) if(i!=Dot) cout<<dig[n][i];
cout<<'e'<<e;

return 0;
}
/*************************************************************End**************************************************************/
__EOF__
数论 模拟
Ural 1238 Folding 题解
Ural 1298 Knight 题解
  • 文章目录
  • 站点概览
Sir_Kay

Sir_Kay

Kaysman #1 Sir_Kay
33 日志
6 分类
34 标签
RSS
Main site Wikipedia GitHub GitLab
Creative Commons
  1. 1. Ural 1248 Sequence Sum 题解
    1. 1.1. 题意
    2. 1.2. 题解
    3. 1.3. 程序
0%
© 2019 – 2020 Sir_Kay