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CodeforcesRound#234(Div.2)

来源:懂视网 责编:小采 时间:2020-11-09 07:39:20
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CodeforcesRound#234(Div.2)

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
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导读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 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;
}

B题:

#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;
}

C题:

#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;
}

D题:

#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;
}

E题:

#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;
}

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文档

CodeforcesRound#234(Div.2)

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
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标签: pro round Codeforces
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