atcoder abc357

A Sanitize Hands

问题：

思路：前缀和，暴力，你想咋做就咋做

代码：

#include <iostream>

using namespace std;

const int N = 2e5 + 10;

int n, m;
int a[N];

int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i ++ ) {
        cin >> a[i];
    }
    
    int ans = 0;
    for(int i = 1; i <= n; i ++ ) {
        m -= a[i];
        ans = i;
        if(m <= 0) break;
    }
    
    if(m < 0) cout << ans - 1;
    else cout << ans;
    return 0;    
}

B Uppercase and Lowercase

问题：

思路：大小写转换，这里有个问题，为什么我的转换最后都变成数字了，先留个疑问

代码：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2e5 + 10;

string str;

int main() {
    cin >> str;
    int cnt1 = 0, cnt2 = 0;
    for(auto t: str) {
        if(t >= 'a' && t <= 'z') cnt1 ++;
        else cnt2 ++;
    }
    
    if(cnt1 >= cnt2)
    transform(str.begin(),str.end(),str.begin(),::tolower);
    else 
    transform(str.begin(),str.end(),str.begin(),::toupper);
    cout<<str<<endl;
    return 0;
}

C Sierpinski carpet

问题：

思路：阴间题，第一眼递归，但是不想求太多坐标，于是想到把图全变成'#'最后填充'.'

代码：

#include <iostream>
#include <cmath>
#include <vector>

using namespace std;

const int N = pow(3, 6) + 10;

char g[N][N];
int n;

int main() {
    cin >> n;
    int len = pow(3, n);
    for(int i = 1; i <= len; i ++ ) {
        for(int j = 1; j <= len; j ++ ) {
            g[i][j] = '#';
        }
    }
    
    for(int level = 1; level <= n; level ++ ) {
        for(int i = 1 + pow(3, level - 1); i <= len; i += pow(3, level)) {
            for(int j = 1 + pow(3, level - 1); j <= len; j += pow(3, level)) {
                for(int k = i; k <= i + pow(3, level - 1) - 1; k ++ ) {
                    for(int u = j; u <= j + pow(3, level - 1) - 1; u ++ ) {
                        g[k][u] = '.';
                    }
                }
            }
        }
    }
    
    for(int i = 1; i <= len; i ++ ) {
        for(int j = 1; j <= len; j ++ ) {
            cout << g[i][j];
        }
        cout << endl;
    }
    return 0;
}

D 88888888

问题：

思路：逆元，快速幂，对原式子变形后发现最后的结果实际上就是x 乘上一个等比数列，这是碰见的第一道逆元的题目，也明确了我对逆元的认识，由于 a / b % mod != (a % mod/ b % mod) % mod，而直接除的话会造成精度丢失，因此我们可以把除法变成乘法，根据费马小定理如果b和p互质，那么b的逆元就等于b ^ p - 2 因此可以快速幂求逆元

代码：

#include <iostream>

using namespace std;

const int mod = 998244353;

long long x;

int get(long long a) {
    int cnt = 0;
    while(a) {
        a /= 10;
        cnt ++;
    }
    return cnt;
}

long long qmi(long long a, long long b) {
    long long res = 1;
    while(b) {
        if(b & 1) res = ((res % mod) * (a % mod)) % mod;
        b >>= 1;
        a = (a % mod * a % mod) % mod;
    }
    return res;
}

int main() {
    cin >> x;
    int len = get(x);
    long long part1 = x % mod;
    long long a = qmi(10, (long long)len);
    long long b = qmi(a, x);
    b --;
    long long c = qmi(a - 1, 998244353 - 2);
    long long part2 = (b % mod * c % mod) % mod;
    cout << (part1 * part2) % mod;
    return 0;
}

E Reachability in Functional Graph

问题：

思路：考虑如果题目是一颗树的话那么直接一个记忆化即可，但是该题会出现环，因此考虑缩点，记得开long long

据说这是基环树板子，回头学一下基环树

代码：

#include <iostream>
#include <cstring>
#include <stack>
#include <map>

using namespace std;

const int N = (2e5 + 10) * 2;

stack<int> stk;
int n;
int val[N], ne[N], h[N], idx;
int dfn[N], low[N], id[N], _size[N], scc_cnt, ts;
int cnt[N];
bool ins[N], st[N];
long long ans = 0;

void add(int a, int b) {
    val[idx] = b;
    ne[idx] = h[a];
    h[a] = idx ++;
}

void tarjan(int u) {
    dfn[u] = low[u] = ++ ts;
    stk.push(u);
    ins[u] = true;
    for(int i = h[u]; i != -1; i = ne[i]) {
        int j = val[i];
        if(!dfn[j]) {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        } else if(ins[j]) low[u] = min(low[u], dfn[j]);
    }
    
    if(dfn[u] == low[u]) {
        ++ scc_cnt;
        int y;
        do {
            y = stk.top();
            stk.pop();
            ins[y] = false;
            id[y] = scc_cnt;
            _size[scc_cnt] ++;
        } while (y != u);
    }
}

void dfs(int u) {
    for(int i = h[u]; i != -1; i = ne[i]) {
        int j = val[i];
        if(!st[j]) {
            dfs(j);
            st[j] = true;
        }
        cnt[u] += cnt[j];
        ans += _size[u] * cnt[j];
    }
}

int main() {
    memset(h, -1, sizeof h);
    cin >> n;
    scc_cnt = n;
    for(int i = 1; i <= n; i ++ ) {
        int x;
        cin >> x;
        add(i, x);
    }
    
    for(int i = 1; i <= n; i ++ ) if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i ++ ) cnt[id[i]] = _size[id[i]];
    map<pair<int, int>, int> ma;
    for(int i = 1; i <= n; i ++ ) {
        for(int j = h[i]; j != -1; j = ne[j]) {
            int k = val[j];
            if(id[i] != id[k] && !ma[{i, k}]) {
                add(id[i], id[k]);
                ma[{i, k}] ++;
            }
        }
    }
    
    memset(st, 0, sizeof st);
    for(int i = scc_cnt; i > n; i -- ) {
        if(!st[i]) {
            st[i] = true;
            dfs(i);
        }
    }
    for(int i = scc_cnt; i > n; i -- ) ans += (long long)_size[i] * (_size[i] - 1);
    cout << ans + n;
    return 0;
}

F two sequence queries

题目：

思路：对sigema a*b做一点变形 设a加上了x，b加上了y

原式 = sigema (a + x) (b + y) = sigema a * b + y * a + x * b + x * y

于是题目变成了区间修改区间查询，显然线段树lazytag板子

代码：这里代码只a了21个数据，应该是哪里没有mod到位或者什么细节没有注意到，短时间内不改了，到期末了

#include <iostream>

using namespace std;

const int N = 2e5 + 10;
const int mod = 998244353;

int n, m;
struct node{
    unsigned long long l, r;
    unsigned long long suma, sumb, sumab;
    unsigned long long taga, tagb;
}tr[4 * N];

void pushup(int u) {
    tr[u].suma = (tr[u << 1].suma + tr[u << 1 | 1].suma) % mod;
    tr[u].sumb = (tr[u << 1].sumb + tr[u << 1 | 1].sumb) % mod;
    tr[u].sumab = (tr[u << 1].sumab + tr[u << 1 | 1].sumab) % mod;
}

void pushdown(int u) {
    tr[u << 1].sumab = (tr[u << 1].sumab + tr[u].taga * tr[u << 1].sumb + tr[u].tagb * tr[u << 1].suma + tr[u].taga * tr[u].tagb * (tr[u << 1].r - tr[u << 1].l + 1)) % mod;
    tr[u << 1 | 1].sumab = (tr[u << 1 | 1].sumab + tr[u].taga * tr[u << 1 | 1].sumb + tr[u].tagb * tr[u << 1 | 1].suma + tr[u].taga * tr[u].tagb * (tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1)) % mod;

    tr[u << 1].suma = (tr[u << 1].suma + (tr[u << 1].r - tr[u << 1].l + 1) * tr[u].taga) % mod;
    tr[u << 1 | 1].suma = (tr[u << 1 | 1].suma + (tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1) * tr[u].taga) % mod;
    tr[u << 1].taga = (tr[u << 1].taga + tr[u].taga) % mod;
    tr[u << 1 | 1].taga = (tr[u << 1 | 1].taga + tr[u].taga) % mod;
    tr[u].taga = 0;
    tr[u << 1].sumb = (tr[u << 1].sumb + (tr[u << 1].r - tr[u << 1].l + 1) * tr[u].tagb) % mod;
    tr[u << 1 | 1].sumb = (tr[u << 1 | 1].sumb + (tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1) * tr[u].tagb) % mod;
    tr[u << 1].tagb = (tr[u << 1].tagb + tr[u].tagb) % mod;
    tr[u << 1 | 1].tagb = (tr[u << 1 | 1].tagb + tr[u].tagb) % mod;
    tr[u].tagb = 0;
}

void build(int u, int l, int r) {
    tr[u].l = l, tr[u].r = r;
    if(l == r) return;
    int mid = l + r >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
}

void add(int u, int p, int x, int type) {
    if(tr[u].l == tr[u].r) {
        if(type == 1) tr[u].suma = x;
        else if(type == 2) tr[u].sumb = x;
        tr[u].sumab = (tr[u].suma * tr[u].sumb) % mod;
    } else {
        int mid = tr[u].l + tr[u].r >> 1;
        if(p <= mid) add(u << 1, p, x, type);
        else add(u << 1 | 1, p, x, type);
        pushup(u);
    }
}

void modify(int u, int l, int r, unsigned long long d, int type) {
    if(tr[u].l >= l && tr[u].r <= r) {
        if(type == 1) {
            tr[u].suma = (tr[u].suma + d * (tr[u].r - tr[u].l + 1)) % mod;
            tr[u].taga = (tr[u].taga + d) % mod;
            tr[u].sumab = (tr[u].sumab + d * tr[u].sumb) % mod;
        } else if(type == 2) {
            tr[u].sumb = (tr[u].sumb + d * (tr[u].r - tr[u].l + 1)) % mod;
            tr[u].tagb = (tr[u].tagb + d) % mod;
            tr[u].sumab = (tr[u].sumab + d * tr[u].suma) % mod;
        }
    } else {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if(mid >= l) modify(u << 1, l, r, d, type);
        if(mid < r) modify(u << 1 | 1, l, r, d, type);
        pushup(u);
    }
}

long long query(int u, int l, int r) {
    if(tr[u].l >= l && tr[u].r <= r) {
        return tr[u].sumab;
    } else {
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        long long res = 0;
        if(mid >= l) res = query(u << 1, l, r);
        if(mid < r) res = (res + query(u << 1 | 1, l, r)) % mod;
        return res;
    }
}

int main() {
    cin >> n >> m;
    build(1, 1, n);
    for(int i = 1; i <= n; i ++ ) {
        int x;
        cin >> x;
        add(1, i, x % mod, 1);
    }    
    for(int i = 1; i <= n; i ++ ) {
        int x;
        cin >> x;
        add(1, i, x % mod, 2);
    }

    while(m -- ) {
        int op;
        cin >> op;
        if(op == 1) {
            int l, r;
            unsigned long long d;
            cin >> l >> r >> d;
            modify(1, l, r, d % mod, 1);
        } else if(op == 2) {
            int l, r, d;
            cin >> l >> r >> d;
            modify(1, l, r, d % mod, 2);
        } else {
            int l, r;
            cin >> l >> r;
            cout << query(1, l, r) << endl;
        }
    }
    return 0;
}