【AtCoder】ARC059

为啥这场ARC那么水……一个点就切完了

ARC059

C - いっしょ / Be Together

枚举就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int64 a[105];

void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    int64 ans = 1e18;
    for(int64 i = -100 ; i <= 100 ; ++i) {
    int64 tmp = 0;
    for(int j = 1 ; j <= N ; ++j) {
        tmp += (a[j] - i) * (a[j] - i);
    }
    ans = min(ans,tmp);
    }
    out(ans);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - アンバランス / Unbalanced

要么是两个相邻的字母相同，要么是两个相同的字母中间隔了一个

其他的状况都会至少包含这两种状况之一

因此只要判这两种就行了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[MAXN];
int N;
void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i < N ; ++i) {
    if(s[i] == s[i + 1]) {out(i);space;out(i + 1);enter;return;}
    if(i != N - 1) {
        if(s[i] == s[i + 2]) {out(i);space;out(i + 2);enter;return;}
    }
    }
    out(-1);space;out(-1);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - キャンディーとN人の子供 / Children and Candies

每一个值其实能够分开考虑

设\(dp[i][j]\)为前i个数指数的总和为j的和，枚举下一个数的指数，计算从\([A_{i + 1},B_{i + 1}]\)的全部状况，更新便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 405
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007; 
int N;
int C,A[MAXN],B[MAXN];
int num[MAXN][MAXN],sum[MAXN][MAXN];
int dp[2][MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(C);
    for(int i = 1 ; i <= N ; ++i) read(A[i]);
    for(int i = 1 ; i <= N ; ++i) read(B[i]);
    for(int i = 1 ; i <= 400 ; ++i) {
    num[i][0] = 1;
    for(int j = 1 ; j <= 400 ; ++j) {
        num[i][j] = mul(num[i][j - 1],i);
    }
    }
    for(int j = 0 ; j <= 400 ; ++j) {
    for(int i = 1 ; i <= 400 ; ++i) {
        sum[i][j] = inc(sum[i - 1][j],num[i][j]);
    }
    }
    int cur = 0;dp[cur][0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
    memset(dp[cur ^ 1],0,sizeof(dp[cur ^ 1]));
    for(int j = 0 ; j <= C ; ++j) {
        for(int h = 0 ; h <= C - j ; ++h) {
        update(dp[cur ^ 1][j + h],mul(dp[cur][j],inc(sum[B[i]][h],MOD - sum[A[i] - 1][h])));
        }
    }
    cur ^= 1;
    }
    out(dp[cur][C]);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - バイナリハック / Unhappy Hacking

这个题就是，我若是新加一个值，默认和我须要的下一位相同，而我摁下一个B后，当前最后一位的值就没必要要是和s中对应位置相同了，因而方案数乘2

设\(dp[i][j]\)为第i次操做，字符串长度为j，这样dp便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
char s[5005];
int dp[5005][5005],N,L;

void Solve() {
    read(N);
    scanf("%s",s + 1);
    L = strlen(s + 1);
    dp[0][0] = 1;
    for(int i = 0 ; i < N ; ++i) {
    for(int j = 0 ; j <= N ; ++j) {
        if(j == 0) update(dp[i + 1][j],dp[i][j]);
        else update(dp[i + 1][j - 1],mul(dp[i][j],2));
        update(dp[i + 1][j + 1],dp[i][j]);
    }
    }
    out(dp[N][L]);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}