2024牛客寒假营Day2||ABEF

原文地址

A-Tokitsukaze and Bracelet

题意

根据手环的三个属性值判断手环的等级。

1. 对攻击百分比来说，+0为100%，+1为150%，+2为200%
2. 对体力和精神来说，+0在 { 29 , 30 , 31 , 32 } \{29,30,31,32\} {29,30,31,32}里选择，+1在 { 34 , 36 , 38 , 40 } \{34,36,38,40\} {34,36,38,40}里选择，+2固定为45

数据范围

n ( 1 ≤ n ≤ 100 ) n(1≤n≤100) n(1≤n≤100)

a i , b i , c i ( a i ∈ { 100 , 150 , 200 } ; b i , c i ∈ { 29 , 30 , 31 , 32 , 34 , 36 , 38 , 40 , 45 } ) a_i,b_i,c_i(a_i∈\{100,150,200\};b_i,ci∈\{29,30,31,32,34,36,38,40,45\}) ai​,bi​,ci​(ai​∈{100,150,200};bi​,ci∈{29,30,31,32,34,36,38,40,45})

思路

模拟即可

参考代码

void solve() {
    int lv1[3] = { 100,150,200 };
    int lv2[9] = { 29,30,31,32,34,36,38,40,45 };
    int a, b, c;cin >> a >> b >> c;
    int ans = 0;
    for (int i = 0;i > m >> k;
    mapcats;
    int ans = 0;
    while (k--) {
        int x, y;cin >> x >> y;
        cats[{ x, y }] = true;
        pairpu = { x - 1,y }, pd = { x + 1,y }, pl = { x,y - 1 }, pr = { x,y + 1 };
        int cnt = 4;
        if (cats.count(pu) != 0)cnt -= 1;
        if (cats.count(pd) != 0)cnt -= 1;
        if (cats.count(pl) != 0)cnt -= 1;
        if (cats.count(pr) != 0)cnt -= 1;
        ans += cnt;
    }
    cout 
    int n;cin  n;
    vectorcol(n + 1);
    setcls;
    for (int i = 1;i 
        cin > col[i];
        cls.insert(col[i]);
    }
    int tn = cls.size();    // 颜色种数
    mapclrs;
    mapcolors;
    int ans = 0;
    int cnt = 0;
    int pi = n;
    while (tn != 0) {
        for (int i = n;i > 0;i--) {
            colors[col[i]]++;
            if (colors[col[i]] == 1) {
                cnt++;  // 达到两次及以上的颜色数
                if (cnt == tn) {
                    ans++;
                    cnt = 0;
                    colors = clrs;
                    pi = i - 1;
                }
            }
        }
        tn = colors.size();
        colors = clrs;
        n = pi;cnt = 0;
    }
    cout