Codeforces Round #234 (Div. 2)

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Problems

#Name A

Inna and Choose Options

standard input/output

1 s, 256 MB

x1942B

Inna and New Matrix of Candies

standard input/output

1 s, 256 MB

x1556C

Inna and Huge Candy Matrix

standard input/output

2 s, 256 MB

x1114D

Dima and Bacteria

standard input/output

2 s, 256 MB

x371E

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<<i)*个数的，个数的计算方式为l * r + l + r。然后去模拟即可。注意初始化的时候要用O(n*b)的方法不然TLE。

代码：

A题：

#include <stdio.h>#include <string.h>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;}

B题：

#include <stdio.h>#include <string.h>#include <algorithm>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;}

C题：

#include <stdio.h>#include <string.h>#include <algorithm>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;}

D题：

#include <stdio.h>#include <string.h>#include <vector>#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;}

E题：

#include <stdio.h>#include <string.h>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;}

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