Codeforces Round #617 (Div. 3)

这场练练golang

题目链接：https://codeforces.com/contest/1296

---

A：

白给

package main

import "fmt"

const (
    maxn = 2010
)

var (
    a    [maxn]int
    n, t int
)

func main() {
    fmt.Scan(&t)
    for cnt := 1; cnt <= t; cnt++ {
        var odd, even int
        fmt.Scan(&n)
        for i := 1; i <= n; i++ {
            var x int
            fmt.Scan(&x)
            if x&1 != 0 {
                odd++
            } else {
                even++
            }
        }
        if odd == 0 {
            fmt.Println("NO")
        } else if even == 0 {
            if n&1 != 0 {
                fmt.Println("YES")
            } else {
                fmt.Println("NO")
            }
        } else {
            fmt.Println("YES")
        }
    }
}

B：

注意每次给钱时保留个位不要给就行

package main

import "fmt"

var (
    t int
)

func main() {
    fmt.Scan(&t)
    for cnt := 1; cnt <= t; cnt++ {
        var s, ans, tmp int
        fmt.Scan(&s)
        tmp = s
        for {
            var refound, yu = tmp / 10, tmp % 10
            ans += refound * 10
            if tmp < 10 {
                ans += tmp
                break
            }
            tmp = refound + yu
        }
        fmt.Println(ans)
    }
}

C：

做偏移量(x,y)到字符串下标的映射即可

package main

import "fmt"

const maxn int = 2e5 + 10

type pair struct {
    fir, sec int
}

var t int

func main() {
    fmt.Scan(&t)
    for ; t > 0; t-- {
        var n int
        var s string
        fmt.Scan(&n, &s)
        s = " " + s
        var ans, l, r, x, y = n + 1, 0, 0, 0, 0
        m := make(map[pair]int)
        m[pair{0, 0}] = 1
        for i := 1; i < len(s); i++ {
            switch s[i] {
            case 'L':
                x--
            case 'R':
                x++
            case 'U':
                y++
            case 'D':
                y--
            }
            v := m[pair{x, y}]
            if v != 0 {
                if ans > (i - v + 1) {
                    ans = i - v + 1
                    l = v
                    r = i
                }
            }
            m[pair{x, y}] = i + 1
        }
        if ans == (n + 1) {
            fmt.Println("-1")
        } else {
            fmt.Println(l, r)
        }
    }
}

D：

对每个数，模a+b之后计算余数要a攻击次数，再O(n)贪心即可

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 2e5 + 10;
int n, a, b, k, ans, h[maxn];

int main() {
    scanf("%d%d%d%d", &n, &a, &b, &k);
    for (int i = 0; i < n; i++) {
        scanf("%d", &h[i]);
        h[i] %= a + b;
        if (!h[i]) h[i] = a + b;
        h[i] = (h[i] - 1) / a;
    }
    sort(h, h + n);
    for (int i = 0; i < n && k >= h[i]; i++) {
        k -= h[i]; ans++;
    }
    printf("%d
", ans);
    return 0;
}