A + B Problem最简单的方法

AC君

2023-11-24 20:40:07

发布于：广东

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这一道题非常简单，话不多说直接上代码：

#include<bits/stdc++.h>
#define str string
using namespace std;

int add(int n)
{
	if (n == 1) return 1;
	return add(n - 1) + 1;
}

int main()
{
	str s, aa, bb;
	int a = 0, b = 2;
	getline(cin, s);
	for (int i = 0; s[i] != ' '; i++)
	{
		a = 0;
		a += b;
		int l[10];
		l[a] += l[b];
		aa += s[i];
	}
	bool fl = false;
	for (int i = 0; i < s.size(); i++)
	{
		if (fl) bb += s[i];
		if (s[i] == ' ')
		{
			fl = true;
		}
	}
	for (int i = 0; i < aa.size(); i++) aa[i] -= 48;
	for (int i = 0; i < bb.size(); i++) bb[i] -= 48;
	int aaa = 0, bbb = 0;
	for (int i = 0; i < aa.size(); i++) aaa = aaa * 10 + aa[i];
	for (int i = 0; i < bb.size(); i++) bbb = bbb * 10 + bb[i];
	int sum = 0, cnt;
	for (int i = 1; i <= aaa; i++)
	{
		sum += i;
		sum -= i - 1;
		int rrr = 0;
		rrr += sum;
		cnt = rrr;
	}
	int sum1 = cnt;
	sum1 += bbb;
	cout << add(sum1);
	return 0;
}

当然，还有更简单的方法：

#define A4C for (int i = 0; s[i] != ' '; i++)
#define A4B if (s[i] == ' ')
#define A4A getline(cin,
#define A49 add(sum1);
#define A48 bb.size();
#define A47 aa.size();
#define A46 s.size();
#define A45 namespace
#define A44 add(int
#define A43 l[10];
#define A42 false;
#define A41 return
#define A40 bb[i];
#define A3F aa[i];
#define A3E main()
#define A3D using
#define A3C aa[i]
#define A3B add(n
#define A3A bb[i]
#define A39 s[i];
#define A38 l[b];
#define A37 true;
#define A36 (fl)
#define A35 (int
#define A34 aaa;
#define A33 i++)
#define A32 cnt;
#define A31 cout
#define A30 bbb;
#define A2F l[a]
#define A2E rrr;
#define A2D std;
#define A2C sum1
#define A2B sum;
#define A2A bool
#define A29 bb;
#define A28 aaa
#define A27 bbb
#define A26 sum
#define A25 int
#define A24 for
#define A23 48;
#define A22 aa,
#define A21 rrr
#define A20 s);
#define A1F str
#define A1E cnt
#define A1D fl
#define A1C bb
#define A1B 2;
#define A1A i;
#define A19 if
#define A18 b;
#define A17 n)
#define A16 (n
#define A15 +=
#define A14 -=
#define A13 0,
#define A12 s,
#define A11 aa
#define A10 0;
#define AF 1)
#define AE 10
#define AD 1;
#define AC <<
#define AB <=
#define AA ==
#define A9 *
#define A8 =
#define A7 b
#define A6 i
#define A5 <
#define A4 -
#define A3 +
#define A2 a
#define A1 {
#define A0 }
#include<bits/stdc++.h>
#define str string
#define A4D A3D A45 A2D A25 A44 A17 A1 A19 A16 AA
#define A4E AF A41 AD A41 A3B A4 AF A3 AD A0
#define A4F A25 A3E A1 A1F A12 A22 A29 A25 A2 A8
#define A50 A13 A7 A8 A1B A4A A20 A4C A1 A2 A8
#define A51 A10 A2 A15 A18 A25 A43 A2F A15 A38 A11
#define A52 A15 A39 A0 A2A A1D A8 A42 A24 A35 A6
#define A53 A8 A10 A6 A5 A46 A33 A1 A19 A36 A1C
#define A54 A15 A39 A4B A1 A1D A8 A37 A0 A0 A24
#define A55 A35 A6 A8 A10 A6 A5 A47 A33 A3C A14
#define A56 A23 A24 A35 A6 A8 A10 A6 A5 A48 A33
#define A57 A3A A14 A23 A25 A28 A8 A13 A27 A8 A10
#define A58 A24 A35 A6 A8 A10 A6 A5 A47 A33 A28
#define A59 A8 A28 A9 AE A3 A3F A24 A35 A6 A8
#define A5A A10 A6 A5 A48 A33 A27 A8 A27 A9 AE
#define A5B A3 A40 A25 A26 A8 A13 A32 A24 A35 A6
#define A5C A8 AD A6 AB A34 A33 A1 A26 A15 A1A
#define A5D A26 A14 A6 A4 AD A25 A21 A8 A10 A21
#define A5E A15 A2B A1E A8 A2E A0 A25 A2C A8 A32
#define A5F A2C A15 A30 A31 AC A49 A41 A10 A0 
#define A60 A4D A4E A4F A50 A51 A52 A53 A54 A55 A56
#define A61 A57 A58 A59 A5A A5B A5C A5D A5E A5F 
#define A62 A60 A61 
#define A63(__FOX__) __FOX__
A63(A62)

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Leo凌

最短：

#include<cstdio>
int main(){printf("5");}

较长：

#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#if __cplusplus >= 201103L
#include<array>
#include<atomic>
#include<chrono>
#include<condition_variable>
#include<forward_list>
#include<future>
#include<initializer_list>
#include<mutex>
#include<random>
#include<ratio>
#include<regex>
#include<scoped_allocator>
#include<system_error>
#include<thread>
#include<tuple>
#include<typeindex>
#include<type_traits>
#include<unordered_map>
#include<unordered_set>
#endif
using namespace std;
int RAND_1(long long a,long long b){
    long long n = a + b;
    cout << n;
    return 0;
}
int RAND_2(string a, string b, int n[500], int m[500], int ans[501]){
    for(int i = 0; i < a.size(); i++)
		n[i] = a[a.size() - i - 1] - '0';
	for(int i = 0; i < b.size(); i++)
		m[i] = b[b.size() - i - 1] - '0';
	int len_maxn = max(a.size(), b.size()) + 1;
	for(int i = 0; i < len_maxn; i++)
	{
		ans[i] = n[i] + m[i] + ans[i];
		if(ans[i] > 9)
		{
			ans[i + 1]++;
			ans[i] = ans[i] % 10;
		}
	}
	while(ans[len_maxn - 1] == 0 && len_maxn > 0) len_maxn--;
	for(int i = len_maxn - 1; i >= 0; i--) cout << ans[i];
	return 0;
}
int n[500], m[500],c[501];
int main(){
    long long a , b;
    int x;
    x = 1 + rand() % 2;
    if(x == 1){
    	cin >> a >> b; 
    	RAND_1(a,b);
	}
    string z , y;
    if(x == 2){
    	cin >> z >> y; 
    	RAND_2(z , y , n , m, c);
	}
    return 0;
}

2025-05-12 来自浙江

氮氧氢氦氩氖氪氙氡氟氯
回复Leo凌
long

//一颗资瓷区间加、区间翻转、区间求和的Splay
#include <bits/stdc++.h>
#define ll long long
#define N 100000
using namespace std;
int sz[N], rev[N], tag[N], sum[N], ch[N][2], fa[N], val[N];
int n, m, rt, x;
void push_up(int x){
sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
sum[x] = sum[ch[x][1]] + sum[ch[x][0]] + val[x];
}
void push_down(int x){
if(rev[x]){
swap(ch[x][0], ch[x][1]);
if(ch[x][1]) rev[ch[x][1]] ^= 1;
if(ch[x][0]) rev[ch[x][0]] ^= 1;
rev[x] = 0;
}
if(tag[x]){
if(ch[x][1]) tag[ch[x][1]] += tag[x], sum[ch[x][1]] += tag[x];
if(ch[x][0]) tag[ch[x][0]] += tag[x], sum[ch[x][0]] += tag[x];
tag[x] = 0;
}
}
void rotate(int x, int &k){
int y = fa[x], z = fa[fa[x]];
int kind = ch[y][1] == x;
if(y == k) k = x;
else ch[z][ch[z][1]==y] = x;
fa[x] = z; fa[y] = x; fa[ch[x][!kind]] = y;
ch[y][kind] = ch[x][!kind]; ch[x][!kind] = y;
push_up(y); push_up(x);
}
void splay(int x, int &k){
while(x != k){
int y = fa[x], z = fa[fa[x]];
if(y != k) if(ch[y][1] == x ^ ch[z][1] == y) rotate(x, k);
else rotate(y, k);
rotate(x, k);
}
}
int kth(int x, int k){
push_down(x);
int r = sz[ch[x][0]]+1;
if(k == r) return x;
if(k < r) return kth(ch[x][0], k);
else return kth(ch[x][1], k-r);
}
void split(int l, int r){
int x = kth(rt, l), y = kth(rt, r+2);
splay(x, rt); splay(y, ch[rt][1]);
}
void rever(int l, int r){
split(l, r);
rev[ch[ch[rt][1]][0]] ^= 1;
}
void add(int l, int r, int v){
split(l, r);
tag[ch[ch[rt][1]][0]] += v;
val[ch[ch[rt][1]][0]] += v;
push_up(ch[ch[rt][1]][0]);
}
int build(int l, int r, int f){
if(l > r) return 0;
if(l == r){
fa[l] = f;
sz[l] = 1;
return l;
}
int mid = l + r >> 1;
ch[mid][0] = build(l, mid-1, mid);
ch[mid][1] = build(mid+1, r, mid);
fa[mid] = f;
push_up(mid);
return mid;
}
int asksum(int l, int r){
split(l, r);
return sum[ch[ch[rt][1]][0]];
}
int main(){
//总共两个数
n = 2;
rt = build(1, n+2, 0);//建树
for(int i = 1; i <= n; i++){
scanf("%d", &x);
add(i, i, x);//区间加
}
rever(1, n);//区间翻转
printf("%d\n", asksum(1, n));//区间求和
return 0;
}

2025-05-23 来自浙江
0

Collapseのbna
还有高手？！

2024-12-29 来自北京
0
李翼飞
《简单》

2023-12-31 来自浙江
0
？
第二个因该是利用宏定义预处理命令在每一个节点展开计算。贪心：你当我不存在？

2023-11-02 来自浙江
0
范特东
官方整活

2023-08-11 来自浙江
0
- 一只姜（AAAAAA级遗址）
  回复范特东
  假官方
  
  2024-04-04 来自浙江
  1
Harry 海云
I think you deliberately want to make simple topics difficult.

2023-07-26 来自广东
0
- Harry 海云
  回复Harry 海云
  When I see this file,I feel very confused
  
  2023-07-26 来自广东
  0

这一道题非常简单，话不多说直接上代码：

当然，还有更简单的方法：

全部评论 6

还有高手？！