模拟进位

M题-传送门

题意：给出一个存储数的方法，给出长度为n的序列v和序列sign，则数值为\(\sum_{i=1}^{n}v_{i}sign_{i}2_{i}(v_{i}\in(0,1)),(sign_{i}\in(-1,1))\)。
现在给sign和序列a，b，求数值a+b的结果c，对应的序列c
思路：
模拟进位
code：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 100;

ll n, a[maxn], b[maxn], c[maxn], sign[maxn];

void work()
{
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> sign[i];
	for(int i = 0; i < n; ++i) cin >> a[i];
	for(int i = 0; i < n; ++i) cin >> b[i];
	ll tmp, pre = 0;
	for(int i = 0; i < n; ++i)
	{
		tmp = sign[i] * (a[i] + b[i]);
		if(sign[i] == 1)
		{
			if(tmp + pre == 3)
				c[i] = 1, pre = 1;
			else if(tmp + pre == 2)
				c[i] = 0, pre = 1;
			else if(tmp + pre == 1)
				c[i] = 1, pre = 0;
			else if(tmp + pre == 0)
				c[i] = 0, pre = 0;
			else if(tmp + pre == -1)
				c[i]++, pre = -1;
		}
		else 
		{
			if(tmp + pre == -3)
				c[i] = 1, pre = -1;
			else if(tmp + pre == -2)
				c[i] = 0, pre = -1;
			else if(tmp + pre == -1)
				c[i] = 1, pre = 0;
			else if(tmp + pre == 0)
				c[i] = 0, pre = 0;
			else if(tmp + pre == 1)
				c[i]++, pre = 1;
		}
	}
	for(int i = 0; i < n; ++i)
	{
		cout << c[i];
		if(i != n - 1) cout << " ";
	}
}

int main()
{
	ios::sync_with_stdio(0);
	work();
	return 0;
}

code2：
可以先求出来10进制下的加和，然后重新模拟填入

#include<bits/stdc++.h>

using namespace std;
using ll = int64_t;

int main() {
    int n; scanf("%d", &n);
    vector<int> sgn(n);
    for (auto &x: sgn)scanf("%d", &x);
    ll a = 0, b = 1;
    for (ll x, i = 0; i < n; ++i) scanf("%lld", &x), a += sgn[i] * (x << i);
    for (ll x, i = 0; i < n; ++i) scanf("%lld", &x), a += sgn[i] * (x << i);
    if (a < 0) b = -1, a = -a;
    for (ll i = 0; i < n; ++i) {
        printf("%d%c", a & 1, " \0"[i == n - 1]);
        a = (a >> 1) + (a & 1 && sgn[i] != b);
    }
    return 0;
}