#ifdef NACHIA
#define _GLIBCXX_DEBUG
#else
#define NDEBUG
#endif
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>
#include <array>
#include <cmath>
// #include <atcoder/modint>
using i64 = long long;
using u64 = unsigned long long;
#define rep(i,n) for(i64 i=0; i<(i64)(n); i++)
#define repr(i,n) for(i64 i=(i64)(n)-1; i>=0; i--)
const i64 INF = 1001001001001001001;
const char* yn(bool x){ return x ? "Yes" : "No"; }
template<typename A> void chmin(A& l, const A& r){ if(r < l) l = r; }
template<typename A> void chmax(A& l, const A& r){ if(l < r) l = r; }
template<typename A> using nega_queue = std::priority_queue<A,std::vector<A>,std::greater<A>>;

#include <iterator>
#include <functional>

using namespace std;

void testcase(){
    i64 N; cin >> N;
    string S; cin >> S;
    vector<i64> C(N); 
    for(auto &i : C) cin >> i;
    vector<i64> Q(N+1), P(N+1);
    rep(i,N) if(i%2 == 1) S[i] ^= '0' ^ '1';
    cout << S << '\n';
    rep(i,N){
        Q[i+1] = Q[i]; P[i+1] = P[i];
        (S[i] == '0' ? Q[i+1] : P[i+1]) += C[i];
    }
    i64 ans = INF;
    for(i64 i=1; i<N; i++) chmin(ans, P[i] + Q[N] - Q[i]);
    for(i64 i=1; i<N; i++) chmin(ans, Q[i] + P[N] - P[i]);
    cout << ans << endl;
}

int main(){
    ios::sync_with_stdio(false); cin.tie(nullptr);
    #ifdef NACHIA
    int T; cin >> T; for(int t=0; t<T; T!=++t?(cout<<'\n'),0:0)
    #endif
    testcase();
    return 0;
}

https://atcoder.jp/contests/abc346/submissions/51551919

蒟蒻昨天晚上想了好久没有想明白，请问一下这份代码的原理是什么？

为什么在进行这一步rep(i,N) if(i%2 == 1) S[i] ^= '0' ^ '1';操作之后，在加一下前缀和，最后就可以了？