Kickstart Round H 2018

打了ks好久都没有更新
诶,自己的粗心真的是没救了,A题大数据都能错
A

#include 
#include 
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#include 
using namespace std;
// const int N = 7005;
// const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

char seq[105][55];
int tot, Root;
int nx[10005][2];
int tag[10005];
int newNode() {
    nx[tot][0] = nx[tot][1] = -1; tag[tot] = 0;
    return tot ++;
}
int N, P; 

void Insert(char* s) {
    int len = strlen(s);
    int root = Root;
    for(int i = 0; i < len; ++i) {
        int id = s[i] == 'R';
        if(nx[root][id] == -1) {
            nx[root][id] = newNode();
        }
        root = nx[root][id];
    }
    tag[root] ++;
}

ll dfs(int rt, int deep) {
    if(tag[rt]) return 1ll<<(N - deep);
    ll ans = 0;
    if(nx[rt][0] != -1) ans += dfs(nx[rt][0], deep + 1);
    if(nx[rt][1] != -1) ans += dfs(nx[rt][1], deep + 1);
    return ans;
}
int main() {
    freopen("A-large-practice2.in", "r", stdin);
    freopen("A-large-practice2.out", "w", stdout);
    int T;
    scanf("%d", &T);
    for(int _ = 1; _ <= T; ++_) {
        tot = 0;
        Root = newNode();
        scanf("%d %d", &N, &P);
        for(int i = 0; i < P; ++i) {
            scanf("%s", seq[i]);
            Insert(seq[i]);
        }
        printf("Case #%d: %lld\n", _, (1ll<

B

#include 
#include 
#include 
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using namespace std;
const int M = 5e6 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

char seq[M];
int main() {
    freopen("B-large.in", "r", stdin);
    freopen("B-large.out", "w", stdout);
    int T;
    scanf("%d", &T);
    for(int _ = 1; _ <= T; ++_) {
        int n;
        scanf("%d %s", &n, seq + 1);
        int ans = 0; int tmp = 0;
        for(int i = 1; i <= (n+1)/2; ++i) {
            tmp += seq[i] - '0';
        }
        ans = max(ans, tmp);
        for(int i = (n+1)/2 + 1, j = 1; i <= n; ++i, ++j) {
            tmp -= seq[j] - '0';
            tmp += seq[i] - '0';
            ans = max(ans, tmp);
        }


        printf("Case #%d: %d\n", _, ans);

    }
    return 0;
}

C容斥原理

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
// const int N = 7005;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

ll Mul[M];
ll Pow(ll x, ll y) {
    ll ans = 1;
    while(y) {
        if(y & 1) ans = 1ll * ans * x % MOD;
        x = 1ll * x * x % MOD;
        y >>= 1;
    }
    return ans;
}

ll C(int x, int y) {
    if(y == 0) return 1;
    if(x == y) return 1;
    return 1ll* Mul[x] * Pow(Mul[y] * Mul[x - y] % MOD, MOD - 2) % MOD;
}
int main() {
    freopen("C-small-attempt0.in", "r", stdin);
    freopen("C-small-attempt0.out2", "w", stdout);
    int T;
    scanf("%d", &T);
    Mul[1] = 1;
    Mul[0] = 1;
    for(int i = 2; i < M; ++i) {
        Mul[i] = Mul[i-1] * i % MOD;
    }
    for(int _ = 1; _ <= T; ++_) {
        int n, m;
        scanf("%d %d", &n, &m);
        ll ans = Mul[2*n];
        int init = 2*n; ll mul2 = 1;
        for(int i = 1; i <= m; ++i) {
            init --;
            mul2 = mul2 * 2 % MOD;
            ll tmp = mul2 * Mul[init] % MOD * C(m, i) % MOD;
            ans = (ans + ( (i % 2) ? -tmp : tmp) + MOD) % MOD;
        }

        printf("Case #%d: %lld\n", _, ans);
    }
    return 0;
}

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