四舍五入1

// Problem: 四舍五入
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1004
// Memory Limit: 204800 MB
// Time Limit: 2000 ms
// 2022-03-16 15:23:10
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
char s[200005];
void solve() {
    int n, t;
    cin >> n >> t;
    cin >> s + 1;
    int idx = -1;
    for (int i = 1 ; i<= n; i ++)
            if (s[i] == '.') {
                idx = i;
                break;
            }
    if (idx == -1) {
        cout << s + 1 << endl;
    }
    else {
       for (int i = idx + 1; i <= n; i ++) {
           if(s[i] >= '5') {
               t --;
               s[i] = '\0';
               i --;
               while (s[i] == '4' && t) {
                   t --;
                   s[i] = '\0';
                   i --;
               }
               if (s[i] == '.') {
                   s[i] = '\0';
                   i --;
                   while (i >= 1 && s[i] == '9') {
                       s[i] = '0';
                       i --;
                   }
                   if (i == 0) cout << 1;
                   else {
                       s[i] += 1;
                   }
               }
               else s[i] += 1;
           }
       }
      cout << s + 1<< endl;
    }
    
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

要求模拟四舍五入，在满足要求的情况下尽可能得到高的分数

可以利用s[i] = '\0'来丢弃一些数