Gym - 101611D Decoding of Varints（边界值处理）

Posted 2021-01-09 yzm10

Decoding of Varints

Statements

Varint is a type used to serializing integers using one or more bytes. The key idea is to have smaller values being encoded with a smaller number of bytes.

First, we would like to encode some unsigned integer x. Consider its binary representation x = a_0a1a2... a_k - 1, where a_i-th stands for the i-th significant bit, i.e. x = a₀·2⁰ + a₁·2¹ + ... + a_k - 1·2^k - 1, while k - 1 stands for the index of the most significant bit set to 1 or k = 1 if x = 0.

To encode x we will use bytes b₀, b₁, ..., b_m - 1. That means one byte for integers from 0 to 127, two bytes for integers from 128 to 2¹⁴ - 1 = 16383 and so on, up to ten bytes for 2⁶⁴ - 1. For bytes b₀, b₁, ..., b_m - 2 the most significant bit is set to 1, while for byte b_m - 1 it is set to 0. Then, for each i from 0 to k - 1, i mod 7 bit of byte is set to a_i. Thus,

x = (b₀ - 128)·2⁰ + (b₁ - 128)·2⁷ + (b₂ - 128)·2¹⁴ + ... + (b_m - 2 - 128)·2^{7·(m - 2)} + b_m - 1·2^{7·(m - 1)}

In the formula above we subtract 128 from b₀, b₁, ..., b_m - 2 because their most significant bit was set to 1.

For example, integer 7 will be represented as a single byte b₀ = 7, while integer 260 is represented as two bytes b₀ = 132 and b₁ = 2.

To represent signed integers we introduce ZigZag encoding. As we want integers of small magnitude to have short representation we map signed integers to unsigned integers as follows. Integer 0 is mapped to 0,  - 1 to 1, 1 to 2,  - 2 to 3, 2 to 4,  - 3 to 5, 3 to 6 and so on, hence the name of the encoding. Formally, if x ≥ 0, it is mapped to 2x, while if x < 0, it is mapped to  - 2x - 1.

For example, integer 75 is mapped to 150 and is encoded as b₀ = 150, b₁ = 1, while  - 75 will be mapped to 149 and will be encoded as b₀ = 149, b₁ = 1. In this problem we only consider such encoding for integers from  - 2⁶³ to 2⁶³ - 1 inclusive.

You are given a sequence of bytes that corresponds to a sequence of signed integers encoded as varints. Your goal is to decode and print the original sequence.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10 000) — the length of the encoded sequence. The next line contains n integers from 0 to 255. You may assume that the input is correct, i.e. there exists a sequence of integers from  - 2⁶³ to 2⁶³ - 1 that is encoded as a sequence of bytes given in the input.

Output

Print the decoded sequence of integers.

Example

Input

5
0 194 31 195 31

Output

0
2017
-2018



首先需要用unsigned long long，除2前值为long long的两倍。
再一个就是先除2再加1，避免先加后值越界。

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const int MAX = 10005;

ll mi[65];
ll a[MAX];

void init(){
    mi[0]=1;
    for(ll i=1;i<=63;i++){
        mi[i]=mi[i-1]*2;
    }
}
int main(void)
{
    int t,i,j;
    ll n,x;
    init();
    scanf("%I64u",&n);
    for(i=1;i<=n;i++){
        scanf("%I64u",&a[i]);
    }
    ll ans=0;int l=0;
    for(i=1;i<=n;i++){
        if(a[i]<128){
            ans+=a[i]*mi[l*7];
            if(ans&1) printf("-%I64u
",ans/2+1);
            else printf("%I64u
",ans/2);

            ans=0;l=0;
            continue;
        }
        ans+=(a[i]-128)*mi[l*7];
        l++;
    }
    return 0;
}