CodeforcesRound#234(Div.2)

Problems # Name A Inna and Choose Options standard input/output 1 s, 256 MB x1942 B Inna and New Matrix of Candies standard input/output 1 s, 256 MB x1556 C Inna and Huge Candy Matrix standard input/output 2 s, 256 MB x1114 D Dima and Bact

Problems

#	Name
A	Inna and Choose Options standard input/output 1 s, 256 MB		x1942
B	Inna and New Matrix of Candies standard input/output 1 s, 256 MB		x1556
C	Inna and Huge Candy Matrix standard input/output 2 s, 256 MB		x1114
D	Dima and Bacteria standard input/output 2 s, 256 MB		x371
E	Inna and Binary Logic standard input/output 3 s, 256 MB		x169

A题：直接暴力枚举每种情况即可。水题

B题：记录下S，G的距离，每种距离只要一次，开个vis数组标记，最后遍历一遍看又几个即可

C题：模拟旋转即可。

D题：并查集+floyd，把w=0的边的点并查集处理，判断同种类是否都在一个集合内，不是就No，剩下的就利用floyd求出最短路即可。

E题：位运算，每个数字对应的每个位向左和向右延生，这个区间内向下的那个三角形区间一定是会增加(1<

代码：

A题：

#include 
#include 

int t, n;
char str[15];
char save[15][15];

bool judge(int a, int b) {
    int i, j;
    for (i = 0; i < a; i++) {
        for (j = 0; j < b; j++)
            save[i][j] = str[i * b + j];
    }
    for (i = 0; i < b; i++) {
        for (j = 0; j < a; j++) {
            if (save[j][i] != 'X') break;
        }
        if (j == a) return true;
    }
    return false;
}

int main() {
    scanf("%d", &t);
    while (t--) {
        scanf("%s", str);
        int ans = 0;
        if (judge(1, 12)) ans++;
        if (judge(2, 6)) ans++;
        if (judge(3, 4)) ans++;
        if (judge(4, 3)) ans++;
        if (judge(6, 2)) ans++;
        if (judge(12, 1)) ans++;
        printf("%d", ans);
        if (judge(1, 12)) printf(" 1x12");
        if (judge(2, 6)) printf(" 2x6");
        if (judge(3, 4)) printf(" 3x4");
        if (judge(4, 3)) printf(" 4x3");
        if (judge(6, 2)) printf(" 6x2");
        if (judge(12, 1)) printf(" 12x1");
        printf("\n");
    }
    return 0;
}

#include 
#include 
#include 
using namespace std;
const int N = 1005;
int n, m, i, j, vis[N];
char g[N][N];

int main() {
    scanf("%d%d", &n, &m);
    for (i = 0; i < n; i++)
        scanf("%s", g[i]);
    for (i = 0; i < n; i++) {
        int G, S;
        for (j = 0; j < m; j++) {
            if (g[i][j] == 'G') G = j;
            if (g[i][j] == 'S') S = j;
        }
        if (S < G) {
            printf("-1\n");
            return 0;
        }
        int d = S - G;
        vis[d] = 1;
    }
    int ans = 0;
    for (int i = 0; i <= 1000; i++)
        if (vis[i]) ans++;
    printf("%d\n", ans);
    return 0;
}

#include 
#include 
#include 
using namespace std;
int n, m, x, y, z, p, i, j;
struct Point {
    int x, y;
} po[100005];

void at(Point &a) {
    int x = a.x, y = a.y;
    a.y = n - x + 1;
    a.x = y;
}

void ht(Point &a) {
    int x = a.x, y = a.y;
    a.y = m - y + 1;
    a.x = x;
}

void ct(Point &a) {
    int x = a.x, y = a.y;
    a.y = x;
    a.x = m - y + 1;
}

int main() {
    scanf("%d%d%d%d%d%d", &n, &m, &x, &y, &z, &p);
    x %= 4;
    y %= 2;
    z %= 4;
    for (i = 0; i < p; i++)
        scanf("%d%d", &po[i].x, &po[i].y);
    for (j = 0; j < x; j++) {
        for (i = 0; i < p; i++) {
            at(po[i]);
        }
        int t = n; n = m; m = t;
    }
    for (j = 0; j < y; j++) {
        for (i = 0; i < p; i++) {
            ht(po[i]);
        }
    }
    for (j = 0; j < z; j++) {
        for (i = 0; i < p; i++) {
            ct(po[i]);
        }
        int t = n; n = m; m = t;
    }
    for (i = 0; i < p; i++)
        printf("%d %d\n", po[i].x, po[i].y);
    return 0;
}

#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
const int N = 100005;
int n, m, K, type[N], i, j, k;
int f[505][505], fa[N], c[505];

int find(int x) {
    if (x == fa[x])
        return x;
    x = find(fa[x]);
}

int main() {
    scanf("%d%d%d", &n, &m, &K);
    int tn = 0;
    memset(f, INF, sizeof(f));
    for (i = 1; i <= n; i++)
        fa[i] = i;
    for (i = 1; i <= K; i++) {
        f[i][i] = 0;
        scanf("%d", &c[i]);
        for (j = 0; j < c[i]; j++) {
            type[++tn] = i;
        }
    }
    int u, v, w;
    while (m--) {
        scanf("%d%d%d", &u, &v, &w);
        if (type[u] != type[v] && f[type[u]][type[v]] > w) {
            f[type[u]][type[v]] = w;
            f[type[v]][type[u]] = w;
        }
        if (w == 0) {
            int pu = find(u);
            int pv = find(v);
            if (pu != pv)
                fa[pv] = pu;
        }
    }
    for (i = 2; i <= n; i++) {
        int u = i - 1, v = i;
        if (type[u] == type[v]) {
            if (find(u) != find(v)) {
                printf("No\n");
                return 0;
            }
        }
    }
    printf("Yes\n");
    for (k = 1; k <= K; k++) {
        for (i = 1; i <= K; i++) {
            for (j = 1; j <= K; j++) {
                f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
            }
        }
    }
    for (i = 1; i <= K; i++) {
        for (j = 1; j < K; j++) {
            if (f[i][j] == INF) f[i][j] = -1;
            printf("%d ", f[i][j]);
        }
        if (f[i][K] == INF) f[i][K] = -1;
        printf("%d\n", f[i][K]);
    }
    return 0;
}

#include 
#include 

const int N = 100005;
const int M = 20;
int n, m, i, j, b;
int a[N][M];
__int64 sum, mi[32];

int main() {
    mi[0] = 1;
    for (i = 1; i < 32; i++)
        mi[i] = mi[i - 1] * 2;
    sum = 0;
    scanf("%d%d", &n, &m);
    int num;
    for (i = 1; i <= n; i++) {
        scanf("%d", &num);
        for (b = 0; b <= 18; b++) {
            if (num&mi[b]) {
                a[i][b] = 1;
            }
        }
    }
    __int64 k = 0;
    for (b = 0; b <= 18; b++) {
        for (i = 1; i <= n; i++) {
            if (a[i][b]) k++;
            else {
                sum += mi[b] * k * (k + 1) / 2;
                k = 0;
            }
        }
        if (k != 0) {
            sum += mi[b] * k * (k + 1) / 2;
                k = 0;
        }
    }
    int p;
    __int64 v;
    while (m--) {
        scanf("%d%I64d", &p, &v);
        __int64 ans1 = 0, ans2 = 0;
        for (b = 0; b <= 18; b++) {
            if (a[p][b]) sum -= mi[b];
            if (a[p][b] && (v&mi[b]) == 0) {
                __int64 l = p, r = p;
                while (a[l - 1][b]) {
                    l--;
                }
                while (a[r + 1][b]) {
                    r++;
                }
                ans1 += mi[b] * ((p - l) * (r - p) + (p - l) + (r - p));
                a[p][b] = 0;
            }
            else if ((v&mi[b]) && a[p][b] == 0) {
                int l = p, r = p;
                while (a[l - 1][b]) {
                    l--;
                }
                while (a[r + 1][b]) {
                    r++;
                }
                ans2 += mi[b] * ((p - l) * (r - p) + (p - l) + (r - p));
                a[p][b] = 1;
            }
        }
        sum = sum - ans1 + ans2 + v;
        printf("%I64d\n", sum);
    }
    return 0;
}