标签：
A：点击打开链接
CSU 1561 (More)Multiplication
题意：把两个数的乘积每个位置的结果填充上去。
注意这个矩阵最大是1e8，所以不能开数组。
模拟题
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 10005;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
int ans[N * 2];
char A[N], B[N];
char a[N * 4];
int main() {
while (~scanf("%s %s", A, B)) {
int m = strlen(A), n = strlen(B), x = 0, y = 0;
for (int i = 0; i < m; i++) {
x = x * 10 + (A[i] - '0');
}
for (int i = 0; i < n; i++) {
y = y * 10 + (B[i] - '0');
}
if (x == 0 && y == 0) break;
int num = x*y, size = 0;
while (num > 0) {
ans[size++] = num % 10;
num /= 10;
}
while (size < n + m) {
ans[size++] = -1;
}
n = n * 4 + 5;
m = m * 4 + 5;
bool f = false;
for (int ii = 0; ii < n; ii++) {
a[m] = '\0';
for (int jj = 0; jj < m; jj++) a[jj] = ' ';
if (ii == 0 || ii == n - 1) {
a[0] = a[m - 1] = '+';
for (int jj = 1; jj < m - 1; jj++) {
a[jj] = '-';
}
}
else {
a[0] = a[m - 1] = '|';
if (ii == 1) {
for (int j = 4; j < m - 1; j += 4) {
a[j] = A[j / 4 - 1];
}
}
else if (ii == n - 2) {
for (int pos = m - 6, id = 0; pos > 0; pos -= 4, id++) {
a[pos] = ans[id] + '0';
a[pos - 2] = '/';
}
if (!f) a[1] = ' ';
}
else {
if ((ii - 2) % 4 == 0) {
for (int j = 2; j < m - 2; j++) {
if ((j - 2) % 4 == 0) a[j] = '+';
else a[j] = '-';
}
}
else {
for (int j = 2; j < m - 2; j += 4) {
a[j] = '|';
}
if ((ii + 1) % 4 == 0) {
if (f) a[1] = '/';
for (int j = 4; j < m - 1; j += 4) {
a[j + 1] = '/';
a[j - 1] = (A[j / 4 - 1] - '0') * (B[(ii + 1) / 4 - 1] - '0') / 10 + '0';
}
}
if (ii % 4 == 0) {
for (int j = 4; j < m - 1; j += 4) {
a[j] = '/';
}
a[m - 2] = B[ii / 4 - 1];
}
if ((ii - 1) % 4 == 0) {
if (ans[size - 1] != -1) {
a[1] = ans[size - 1] + '0';
f = true;
}
size--;
for (int j = 4; j < m - 1; j += 4) {
a[j - 1] = '/';
a[j + 1] = (A[j / 4 - 1] - '0') * (B[(ii - 1) / 4 - 1] - '0') % 10 + '0';
}
}
}
}
}
printf("%s\n", a);
}
}
return 0;
}B:点击打开链接题意：
给定一个矩阵，矩阵内部有2种镜子/ \，从边缘的*入射一条光线，则光线一定会射到边缘，把边缘用&标记。
然后输出这个矩阵。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
const int MAX_N = 27;
int n, m;
int ex, ey;
char maz[27][27];
bool vis[MAX_N][MAX_N];
bool found;
bool safe(int x, int y) {
return x >= 0 && x < m && y >= 0 && y < n;
}
void dfs(int x, int y, int dir) {
if (found) return ;
if (dir == 1) ++x;
else if (dir == 2) ++y;
else if (dir == 3) --x;
else --y;
if (safe(x, y)) {
if (maz[x][y] == 'x') {
ex = x, ey = y;
found = true;
return ;
}
if (maz[x][y] == '.') dfs(x, y, dir);
else if (maz[x][y] == '/') {
if (dir == 1) {
dfs(x, y, 4);
} else if (dir == 2) {
dfs(x, y, 3);
} else if (dir == 3) {
dfs(x, y, 2);
} else {
dfs(x, y, 1);
}
} else {
if (dir == 1) {
dfs(x, y, 2);
} else if (dir == 2) {
dfs(x, y, 1);
} else if (dir == 3) {
dfs(x, y, 4);
} else {
dfs(x, y, 3);
}
}
}
}
void Clear() {
ex = 0, ey = 0;
found = false;
memset(maz, 0, sizeof maz);
}
int main() {
int cas = 0;
while (2 == scanf("%d%d", &n, &m)) {
if (0 == n && 0 == m) break;
int sx = 0, sy = 0;
Clear();
for (int i = 0; i < m; ++i) {
scanf("%s", maz[i]);
for (int j = 0; j < n; ++j) {
if (maz[i][j] == '*') {
sx = i, sy = j;
maz[i][j] = '.';
}
}
}
if (sx == 0)
dfs(sx, sy, 1);
else if (sx == m-1)
dfs(sx, sy, 3);
else if (sy == 0)
dfs(sx, sy, 2);
else
dfs(sx, sy, 4);
printf("HOUSE %d\n", ++cas);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (i == sx && j == sy) {
putchar('*');
} else if (i == ex && j == ey) {
putchar('&');
} else putchar(maz[i][j]);
}
puts("");
}
}
return 0;
}C:点击打开链接给定一个字符串，问字典序第k小的是什么。
从高位到低位枚举，每次填充排列数<=k且最大的那个字母。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 20;
char s[N], t[N];
ll k;
ll num[27];
ll J[30];
int len;
ll way(){
ll sum = 0;
for(int i = 0; i < 26; i++)sum += num[i];
ll ans = J[sum];
for(int i= 0; i < 26; i++)ans /= J[num[i]];
return ans;
}
void dfs(int now , ll pre){
if(now == len)return ;
char c = 'A';
for(int i = 0; i < 26; i++){
if(num[i]==0)continue;
c = 'A'+i;
num[i]--;
ll hehe = way();
num[i]++;
if(pre + hehe >= k){
break;
}
pre += hehe;
}
num[c-'A']--;
t[now] = c;
dfs(now+1, pre);
}
int main() {
J[0] = 1;
for(ll i = 1; i < 30; i++)J[i] = J[i-1]*i;
while (~scanf("%s", s)) {
cin>>k;
if(s[0] == '#' && k==0)break;
memset(num, 0, sizeof num);
len = strlen(s);
memset(t, 0, sizeof t);
for(int i = 0; s[i]; i++)num[s[i]-'A'] ++;
dfs(0, 0);
puts(t);
}
return 0;
}CSU 1565 Word Cloud
某模拟题。
给出矩阵的宽度为P， N个字母。
下面给出每个字母和它的高度。
用题目中的图求出这个字母的宽度。
把字母从上到下，从左到右依次填充进矩阵，问填充完后矩阵的高度是多少
注意字母间距离是 10px
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
const int MAX_N = 107;
int w, n;
char str[100007];
int c[MAX_N], len[MAX_N];
int H[MAX_N], W[MAX_N];
int main() {
int cas = 0;
while (2 == scanf("%d%d", &w, &n)) {
if (0 == w && 0 == n) break;
int Cmax = 0;
for (int i = 0; i < n; ++i) {
scanf("%s%d", str, &c[i]);
Cmax = max(Cmax, c[i]);
len[i] = strlen(str);
}
for (int i = 0; i < n; ++i) {
H[i] = 8 + ceil(40 * (c[i] - 4) * 1. / (Cmax - 4));
W[i] = ceil(9. * len[i] * H[i] / 16);
}
long long ans = 0, sum = 0;
bool row = false;
int maxH = 0;
w += 10;
for (int i = 0; i < n; ++i) {
if (row) {
row = false;
ans += maxH;
sum = 0;
maxH = 0;
}
if (sum + W[i] + 10 <= w) {
maxH = max(maxH, H[i]);
sum += W[i] + 10;
} else row = true, --i;
}
if (sum != 0) ans += maxH;
printf("CLOUD %d: %lld\n", ++cas, ans);
}
return 0;
}F:点击打开链接给定一个迷宫的图，每个格子有16种形态，用一个16进制数表示。
边缘有且仅有2个缺口。
问：
if 2个缺口不连通则输出"No solution"
else if 存在点是不可达的则输出“Unreachable cell"
else if 2个缺口间存在不同的路径则输出 "Multiple path"
else "Maze ok"
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 55;
int n, m;
vector<int>G[N*N], path;
vector<pii>D;
int has(int x, int y){return x*m+y;}
void add(int x, int y, int i, int j){
G[has(x,y)].push_back(has(i,j));
}
int getx(int val){return val/m;}
int gety(int val){return val%m;}
bool inmap(int x, int y){
return 0<=x && x < n && 0<=y&&y<m;
}
char s[N];
int mp[N][N];
int step[4][2] = {0,-1, 1,0, 0,1, -1,0};
bool vis[N][N];
void bfs(int x, int y){
memset(vis, 0, sizeof vis);
queue<int>q;
q.push(has(x,y));
vis[x][y] = true;
while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].size(); i++){
int v = G[u][i];
x = getx(v); y = gety(v);
if(vis[x][y])continue;
vis[x][y] = true;
q.push(v);
}
}
}
int solve(){
if(false == vis[D[1].first][D[1].second])return 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(false == vis[i][j])return 1;
int siz = 0;
for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)siz += G[has(i,j)].size();
if((siz>>1)==n*m-1)return 3;
return 2;
}
int main() {
while (~scanf("%d %d", &n, &m), n+m) {
D.clear();
for(int i = 0; i < n; i++){
scanf("%s", s);
for(int j = 0; j < m; j++)
{
G[has(i,j)].clear();
if('0' <= s[j] && s[j]<='9')mp[i][j] = s[j]-'0';
else mp[i][j] = s[j]-'A'+10;
}
}
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
for(int k = 0; k < 4; k++)
{
int x = i+step[k][0], y = j+step[k][1];
if(!inmap(x,y)){
if((mp[i][j]&(1<<k))==0)
D.push_back(pii(i,j));
continue;
}
if((mp[i][j]&(1<<k))==0)
add(i,j, x,y);
}
sort(D.begin(), D.end());
D.erase(unique(D.begin(), D.end()), D.end());
if((int)D.size()==1) D.push_back(D[0]);
bfs(D[0].first, D[0].second);
int ans = solve();
if(ans == 0)puts("NO SOLUTION");
else if(ans == 1) puts("UNREACHABLE CELL");
else if(ans == 2)puts("MULTIPLE PATHS");
else puts("MAZE OK");
}
return 0;
}
/*
1 8
2AAAAAAE
*/G:点击打开链接#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
const int MAX_N = 57;
const char table[30] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ_.";
int n;
char str[MAX_N];
int ID(char ch) {
for (int i = 0; table[i]; ++i) if (table[i] == ch) return i;
}
int main() {
while (1 == scanf("%d", &n)) {
if (0 == n) break;
scanf("%s", str);
for (int i = 0; str[i]; ++i) {
str[i] = table[(ID(str[i]) + n) % 28];
}
int len = strlen(str);
for (int i = len - 1; i >= 0; --i) {
putchar(str[i]);
}
puts("");
}
return 0;
}H:点击打开链接
CSU 1568 Shrine Maintenance
题意：输入M N, D 后面D个数
把一个半径为1000的圆N等分，再利用圆上N个点中的所有有效点画出恰好M个类三角形（我想说的是一个类三角形和圆心相连的边恰好有两条哟，即使第二图中只有一个点相连）
有效点：我们给圆中的点编号(1~N) ，那么是D个数中任意一个数的倍数 就是有效点。
问：
使得画出的M个类三角形中 最大的周长最小。 输出那个周长， 保留一位小数。
思路：
二分答案判可行。
我们把相邻的有效点两两相连
W的作用相当于是 断开了W条有效边
假设W=0，那么应该是所有边长相连，所以是一个环。
开始的做法是枚举环的第一个断点，但这样会t
而我们发现前面的因子间隔比较大，达到lcm后间隔就相同,这样断开任意一边都是等效的。也就是说枚举前面的因子即可。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
const int MAX_N = 8607;
int W, N, D;
int cnt;
bool vis[MAX_N];
int id[MAX_N];
void Clear() {
cnt = 0;
memset(vis, false, sizeof vis);
}
double get(int i, int j) {
return 2000. * sin(acos(-1.) * abs(id[j] - id[i]) / N);
}
bool check(double ans, int j) {
int w = W;
double sum = 0;
for (int i = (j+2)%cnt; ; i = (i+1)%cnt) {
double now = get((i - 1+cnt)%cnt, i);
if (sum + now <= ans) {
sum += now;
}
else {
sum = 0;
--w;
if (w < 0) return false;
}
if(i==j)break;
}
return true;
}
int main() {
while (1 == scanf("%d", &W)) {
if (0 == W) break;
Clear();
scanf("%d%d", &N, &D);
for (int i = 0; i < D; ++i) {
int v;
scanf("%d", &v);
for (int j = v; j <= N; j += v)
vis[j] = true;
}
for (int i = 1; i <= N; ++i) if (vis[i])
id[cnt++] = i;
if(cnt <= W) {
puts("2000.0");
continue;
}
if(W == 0){
puts("0.0"); continue;
}
W--;
double ans = 1e8;
for(int j = 0; j < min(100,cnt); j++){
double l = 0, r = ans;
for (int i = 0; i <30; ++i) {
double mid = (l + r) / 2;
if (check(mid, j)) r = mid;
else l = mid;
}
ans = min(ans, r);
}
printf("%.1f\n", ans + 2000);
}
return 0;
}CSU 1569 Wet Tiles
给定n*m的矩阵 时刻t L个起点 W个墙
下面给出L个点的坐标
下面每4个数字给出每个墙的起点终点（墙要么平行于轴，要么45°倾斜）
问每个点bfs出去t下后，整个图被遍历了多少点。
简单bfs
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
typedef pair<int, int> pii;
const int N = 1005;
char mp[N][N];
int vis[N][N];
int n, m, T;
queue<pii> que;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {-1, 1, 0, 0};
int bfs() {
int ans = 0;
while(que.size() > 0) {
pii tmp = que.front(); que.pop();
int x = tmp.first, y = tmp.second;
ans ++;
if(vis[x][y] == T) continue;
for(int i = 0; i < 4; i ++) {
int nx = x + dx[i];
int ny = y + dy[i];
if(nx >= 0 && nx < n && ny >= 0 && ny < m && vis[nx][ny] == 0) {
vis[nx][ny] = vis[x][y] + 1;
que.push(make_pair(nx, ny));
}
}
}
return ans;
}
int main() {
while(~scanf("%d", &m)) {
if(m == -1) break;
memset(vis, 0, sizeof vis);
int L, W;
scanf("%d%d%d%d", &n, &T, &L, &W);
for(int i = 0, x, y; i < L; i ++) {
scanf("%d%d", &x, &y);
swap(x, y);
y--, x = n - x;
que.push(make_pair(x, y));
vis[x][y] = 1;
}
for(int i = 0, x1, x2, y1, y2; i < W; i ++) {
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
swap(x1, y1); swap(x2, y2);
y1--, y2--;
x1 = n - x1;
x2 = n - x2;
if(x1 == x2) {
if(y1 < y2) {
while(y1 <= y2) {
vis[x1][y1++] = -1;
}
} else {
while(y1 >= y2) {
vis[x1][y1--] = -1;
}
}
} else if(y1 == y2) {
if(x1 < x2) {
while(x1 <= x2) {
vis[x1++][y1] = -1;
}
} else {
while(x1 >= x2) {
vis[x1--][y1] = -1;
}
}
} else {
if(x1 < x2) {
if(y1 < y2) {
while(x1 <= x2) {
vis[x1++][y1++] = -1;
}
} else {
while(x1 <= x2) {
vis[x1++][y1--] = -1;
}
}
} else {
if(y1 < y2) {
while(x1 >= x2) {
vis[x1--][y1++] = -1;
}
} else {
while(x1 >= x2) {
vis[x1--][y1--] = -1;
}
}
}
}
}
printf("%d\n", bfs()); }
return 0;
}
湖南多校对抗赛（2015.4.6）CSU 1561~1569 题解
标签：
原文地址：http://blog.csdn.net/qq574857122/article/details/45128101
