C. Rings（构造）

https://codeforces.com/problemset/problem/1562/C
题意
Frodo was caught by Saruman. He tore a pouch from Frodo’s neck, shook out its contents —there was a pile of different rings: gold and silver…
“How am I to tell which is the One?!” the mage howled.

“Throw them one by one into the Cracks of Doom and watch when Mordor falls!”

Somewhere in a parallel Middle-earth, when Saruman caught Frodo, he only found n rings. And the i-th ring was either gold or silver. For convenience Saruman wrote down a binary string s of n characters, where the i-th character was 0 if the i-th ring was gold, and 1 if it was silver.

Saruman has a magic function f, which takes a binary string and returns a number obtained by converting the string into a binary number and then converting the binary number into a decimal number. For example, f(001010)=10,f(111)=7,f(11011101)=221.

Saruman, however, thinks that the order of the rings plays some important role. He wants to find 2 pairs of integers (l1,r1),(l2,r2), such that:

1≤l1≤n, 1≤r1≤n, r1−l1+1≥⌊n2⌋
1≤l2≤n, 1≤r2≤n, r2−l2+1≥⌊n2⌋
Pairs (l1,r1) and (l2,r2) are distinct. That is, at least one of l1≠l2 and r1≠r2 must hold.
Let t be the substring s[l1:r1] of s, and w be the substring s[l2:r2] of s. Then there exists non-negative integer k, such that f(t)=f(w)⋅k.
Here substring s[l:r] denotes slsl+1…sr−1sr, and ⌊x⌋ denotes rounding the number down to the nearest integer.

Help Saruman solve this problem! It is guaranteed that under the constraints of the problem at least one solution exists.

input
Each test contains multiple test cases.

The first line contains one positive integer t (1≤t≤103), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer n (2≤n≤2⋅104) — length of the string.

The second line of each test case contains a non-empty binary string of length n.

It is guaranteed that the sum of n over all test cases does not exceed 105.

output
For every test case print four integers l1, r1, l2, r2, which denote the beginning of the first substring, the end of the first substring, the beginning of the second substring, and the end of the second substring, respectively.

If there are multiple solutions, print any.

思路
本题暴力必超时，我们考虑构造，我们发现对于一个二进制数，在它后面增加一个0便是将这个数乘2，例如110（6）的两倍1100（12），111（7）的两倍1110（14），所以我们找01串中0的位置，当0出现在前n/2个数中时，我们考虑向后构造t和w相等的情况，当0出现在后n/2个数中时，我们考虑向前构造t是w两倍的情况，当没0时在n/2附件向前向后拓展成相等的串即可。

代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
char s[N];
int main()
{int t;cin>>t;while(t--){int n;cin>>n>>s+1;int pos1=-1,pos2=-1;for(int i=1;s[i];i++){if(i>n/2&&s[i]=='0') pos2=i;else if(i<=n/2&&s[i]=='0') pos1=i;}if(pos2!=-1) cout<<"1 "<<pos2<<" 1 "<<pos2-1<<endl;else if(pos1!=-1) cout<<pos1<<" "<<n<<" "<<pos1+1<<" "<<n<<endl;else if(n&1) cout<<"1 "<<n/2+1<<" "<<n/2+1<<" "<<n<<endl;else cout<<"1 "<<n/2<<" "<<n/2+1<<" "<<n<<endl;}return 0;
}

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C. Rings（构造）

C. Rings（构造）

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