Codeforces Round #487 (Div. 2)

A. A Blend of Springtime(暴力/模拟)

题目大意

给出$n$个花，每个点都有自己的颜色，问是否存在连续大于等于三个花颜色均不相同

sol

直接模拟判断即可

#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN  = 1001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int a[MAXN][4];
char s[MAXN];
int main() {
    #ifdef WIN32
    //freopen("a.in", "r", stdin);
    #endif
    scanf("%s", s + 1);
    int N = strlen(s + 1);
    for(int i = 1; i <= N; i++) {
        if(s[i] == 'A') a[i - 1][1] = 1, a[i + 1][1] = 1, a[i][1] = 1;
        if(s[i] == 'B') a[i - 1][2] = 1, a[i + 1][2] = 1, a[i][2] = 1;
        if(s[i] == 'C') a[i - 1][3] = 1, a[i + 1][3] = 1, a[i][3] = 1;
    }
    for(int i = 1; i <= N; i++) {
        if(a[i][1] == 1 && a[i][2] == 1 && a[i][3] == 1) {
            puts("Yes"); return 0;
        }
    }
    puts("No");
    return 0;
}

B. A Tide of Riverscape(暴力/模拟)

题目大意

给定一段序列，由$“1”,“0”,“.”$组成，其中$.$代表不确定是$“1”$还是$“0”$，

给定一个$p$，问这个序列是否满足对于$i + P <= N$的$i$，存在$i$与$i+P$位置的字符不同。

sol

大力特判两个位置是否可以满足

#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN  = 2001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, P;
char s[MAXN];
#define GG puts("No"); return 0;
int main() {
    #ifdef WIN32
    //freopen("a.in", "r", stdin);
    #endif
    scanf("%d %d", &N, &P);
    scanf("%s", s + 1);
    bool flag = 0;
    for(int i = 1; i <= N - P; i++) {
        if(s[i] == '1') {
            if(s[i + P] == '0') {flag = 1; break;}
            if(s[i + P] == '.') {s[i + P] = '0'; flag = 1; break;}
        }
        if(s[i] == '0') {
            if(s[i + P] == '1') {flag = 1; break;}
            if(s[i + P] == '.') {s[i + P] = '1'; flag = 1; break;}
        }
        if(s[i] == '.') {
            if(s[i + P] == '1') {s[i] = '0'; flag = 1; break;}
            if(s[i + P] == '0') {s[i] = '1'; flag = 1; break;}
            if(s[i + P] == '.') {s[i] = '1'; s[i + P] = '0'; flag = 1; break;}
        }
    }
    if(flag == 0) {puts("No"); return 0;}
    for(int i = 1; i <= N; i++) {
        if(s[i] == '.') putchar('0');
        else putchar(s[i]);
    }
    return 0;
}

C. A Mist of Florescence(构造)

题目大意

给出四个数$a,b,c,d$，构造一个矩阵满足$“A”,"B","C","D"$对应联通块的数量为$a,b,c,d$

sol

考场上没想出来，思维太局限了，看到$n,m<=50$但是没有把它作为突破口。

正解非常刁钻，一图解千愁，不过我写的和正解不太一样，我是每三个空格放一个。

#include<cstdio>
using namespace std;
const int MAXN = 51;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int mp[MAXN][MAXN];
int color[MAXN] = {0, 0, 1, 2, 3};
char ans[MAXN] = {' ', 'A', 'B', 'C', 'D'};
int a[5];
int main() {
    #ifdef WIN32
    //freopen("a.in", "r", stdin);
    #endif
    //int a = read() - 1, b = read() - 1, c = read() - 1, d = read() - 1;
    for(int i = 1; i <= 4; i++) a[i] = read() - 1;
    for(int i = 1; i <= 4; i++) 
        for(int k = color[i] * 12 + 1; k <= color[i] * 12 + 13; k++)
            for(int j = 1; j <= 50; j++) 
                mp[k][j] = color[i] + 1;
    
    /* for(int i = 1; i <= MAXN - 1; i++, puts(""))
        for(int j = 1; j <= MAXN - 1; j++)
            printf("%d ", mp[i][j]); */
    for(int i = 1; i <= 4; i++) {
        int num = a[i];
        for(int k = color[5 - i] * 12 + 2; num > 0 && k <= color[5 - i] * 12 + 12; k++) {
            for(int j = 2 + (k & 1); num > 0 && j <= 49; j += 3)
                mp[k][j] = color[i] + 1, num--;
        }
    }
    printf("48 50
");
    for(int i = 1; i <= 48; i++, puts(""))
        for(int j = 1; j <= 50; j++)
            putchar(ans[mp[i][j]]);
    return 0;
}

总结

又是两题滚粗，不过值得庆幸的是前两题都是1A，T3没做出来确实比较遗憾

以前从来没做过构造题也是原因之一

感觉T3这种题是有点套路的，最重要的是不要相信它给的样例！！！

然后应该把思维打开，多在宏观角度构造构造。