# [algorithm learning] sword finger offer 64 Find 1 + 2 +... + n (Java / C / C + + / Python / go / trust)

2022-01-29 14:07:01

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# The finger of the sword Offer 64. seek 1+2+…+n：

seek 1+2+...+n , It is required that multiplication and division shall not be used 、for、while、if、else、switch、case Wait for keywords and conditional statements （A?B:C）.

# Examples 1

`````` Input :
n = 3
Output :
6
Copy code ``````

# Examples 2

`````` Input :
n = 9
Output :
45
Copy code ``````

# Limit

• 1 <= n <= 10000

# analysis

• The usual practice is to multiply , loop , recursive .
• The problem does not allow multiplication and circulation , Then use recursion instead of ; Recursion requires a termination condition , The problem is not judged by conditions , We can use short circuit logic operators instead of .
• Recursion is too slow , We can use the sum formula of the sequence of equal differences , It becomes a calculation n * (n + 1) / 2.
• Divide 2 You can replace... With a shift to the right .
• Multiplication , We can consider the steps of multiplication in primary school ,a × b, We just use b Multiply each digit of by a × 10^i^,i Express b The number of digits of , Add up the results . Then if b yes 2 Every bit of the base , It becomes watching b Every one of you is 0 still 1, If it is 1 Just add 1 × a × 2^i^, Replace it with a shift and it becomes a << i.
• Within limits n The biggest is 10000,2^14^ That's enough .

## java

``````class Solution {
public int sumNums(int n) {
// 2 Of 14 More than... Times 10000
//  The summation formula of the sequence of equal differences n * (n + 1) / 2
int ans = 0, a = n, b = n + 1;
//  Not much use for eggs. , In order to turn an expression into a statement
boolean flag;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;
a <<= 1;
b >>= 1;

flag = ((b & 1) > 0) && (ans += a) > 0;

return ans >> 1;
}
}
Copy code ``````

## c

``````int sumNums(int n){
// 2 Of 14 More than... Times 10000
//  The summation formula of the sequence of equal differences n * (n + 1) / 2
int ans = 0, a = n, b = n + 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);

return ans >> 1;
}
Copy code ``````

## c++

``````class Solution {
public:
int sumNums(int n) {
// 2 Of 14 More than... Times 10000
//  The summation formula of the sequence of equal differences n * (n + 1) / 2
int ans = 0, a = n, b = n + 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);
a <<= 1;
b >>= 1;

(b & 1) && (ans += a);

return ans >> 1;
}
};
Copy code ``````

## python

``````class Solution:
def sumNums(self, n: int) -> int:
# 2 Of 14 More than... Times 10000
#  The summation formula of the sequence of equal differences n * (n + 1) / 2
ans = 0
a = n
b = n + 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)
a <<= 1
b >>= 1

(b & 1) and (ans := ans + a)

return ans >> 1
Copy code ``````

## go

``````func sumNums(n int) int {
// 2 Of 14 More than... Times 10000
//  The summation formula of the sequence of equal differences n * (n + 1) / 2
ans, a, b := 0, n, n + 1
addGreatZero := func() bool {
ans += a
return ans > 0
}

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()
a <<= 1
b >>= 1

_ = ((b & 1) > 0) && addGreatZero()

return ans >> 1
}
Copy code ``````

## rust

``````impl Solution {
pub fn sum_nums(n: i32) -> i32 {
let mut ans = 0;
let mut a = n;
let mut b = n + 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);
a <<= 1;
b >>= 1;

((b & 1) > 0) && Solution::add_great_zero(&mut ans, a);

ans >> 1
}

fn add_great_zero(ans: &mut i32, n: i32) -> bool {
*ans += n;
*ans > 0
}
}
Copy code `````` 