继续搞起,慢慢来,不着急~
代码如下:
Uva10935,丢掉卡片
#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
deque<int>p, pp;
int main() {
int n;
while (scanf("%d", &n)==1 && n != 0) {
p.clear(), pp.clear();
for (int i = 1; i <= n; i++)
p.push_back(i);
while (p.size() != 1) {
pp.push_back(p.front()), p.pop_front();
p.push_back(p.front()), p.pop_front();
}
printf("Discarded cards:");
for (int i = 0; i < pp.size(); i++) {
if (i == pp.size() - 1)
printf(" %d", pp[i]);
else
printf(" %d,", pp[i]);
}
printf("
Remaining card: %d
", p[0]);
}
return 0;
}
Uva1595,对称
#include<cstdio>
#include<iostream>
#include<map>
#include<cmath>
#include<utility>
using namespace std;
typedef multimap<int, int> Map;
Map p, pp;
int main() {
//FILE*fp = fopen("stdout.txt", "w");
int kase = 0, n;
scanf("%d", &n);
while (kase++ < n) {
p.clear(), pp.clear();
int num, a, b;
scanf("%d", &num);
for (int i = 0; i < num; i++) {
scanf("%d %d", &a, &b);
p.insert(make_pair(a, b));
}
Map::iterator it = p.begin(), itt = --p.end(), ins = p.begin(), tar, mid;
if (num % 2 == 1) {
advance(ins, ceil(num / 2));
p.insert(*ins);
}
int sum = it->first + itt->first, flag = 1;
for (advance(it,p.size()/2); it != p.end(); it++) {
pp.insert(*it);
}
int tot = p.size();
it = p.begin(), itt = --p.end();
if (num == 1|| it->first == itt->first) {
//fprintf(fp,"YES
");
printf("YES
");
continue;
}
for (int i = 0; i < p.size() / 2; i++) {
flag = 0;
for (itt = pp.begin(); itt != pp.end(); itt++) {
if (it->first == itt->first) {
flag = 1;
pp.erase(itt);
break;
}
if (itt->first + it->first == sum&&itt->second == it->second) {
flag = 1;
pp.erase(itt);
break;
}
}
if (!flag)
break;
it++;
}
if (flag)
//fprintf(fp,"YES
");
printf("YES
");
else
//fprintf(fp,"NO
");
printf("NO
");
}
return 0;
}
Uva12100,打印队列
#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
int main() {
int kase;
scanf("%d", &kase);
while (kase--) {
queue<int>p;
int str[10] = { 0,0,0,0,0,0,0,0,0,0 };
int n, m, mid, time = 0, pos, level;
scanf("%d %d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%d", &mid);
p.push(mid);
str[mid]++;
if (i == m) {
pos = str[mid];
level = mid;
}
}
bool flag = 1;
for (int i = 9; i > level; i--) {
while (str[i] != 0) {
if (p.front() != i) {
if (p.front() == level) {
if (pos != 1)
pos--;
else
pos = str[level];
}
p.push(p.front()), p.pop();
}
else {
p.pop();
str[i]--;
time++;
}
}
}
printf("%d
", time + pos);
}
return 0;
}
算法课第一周作业,颠倒的密码
#include<cstdio>
#include<iostream>
#include<cmath>
int main() {
int a[100];
for (int n = 1000; n <= 9999; n++) {
int ini = n, i = 0, new_num = 0;
do {
a[i++] = ini % 10;
ini /= 10;
} while (ini);
int str = i;
for (int j = 0; j < str; j++, i--) {
new_num += a[j] * pow(10, i - 1);
}
if (new_num == n * 4) {
printf("%d", new_num);
break;
}
}
return 0;
}
算法课第二周作业,相亲数(素数表打印,数组下表的嵌套)
#include<cstdio>
#include<iostream>
using namespace std;
const int max = 30000;
long table[max];
long flag[max];
int main() {
for (long i = 2, j = 0; i < max; i++) {
j = (i << 1);
while (j < max) {
table[j] += i;
j += i;
}
}
for (long i = 2; i < max; i++) {
if (flag[i] != 0) {
if (table[i] < max&&(table[table[i] + 1] + 1 == i))
printf("%ld and %ld
", i, table[i] + 1);
flag[table[i] + 1] = 1;
}
}
return 0;
}
算法课第三周作业,高精度除法
#include<cstdio>
#include<iostream>
using namespace std;
const int num = 30;//要求除到的精度
int main() {
int dividend, divisor, mid;//被除数,除数,中间变量
int str[100] = {};//储存数组
scanf("%d%d", ÷nd, &divisor);
for (int i = 0; i < num; i++) {
mid = dividend / divisor;
if (mid != 0)
dividend = 10 * (dividend % divisor);
else
dividend *= 10;
str[i] = mid;
}
printf("%d.", str[0]);
for (int i = 1; i < num; i++)
printf("%d", str[i]);
return 0;
}
Uva1596,找bug
#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<utility>
using namespace std;
long long arrlen[100];
vector<map<string,string>>arrcnt[100];
void clear() {
map<string, string>a;
for (int i = 0; i < 58; i++) {
arrlen[i] = 0;
arrcnt[i].clear();
arrcnt[i].push_back(a);
}
}
int de_bug(string s, string ss) {
int pos1, pos2, end1, end2;
string mid, mid1;
pos1 = s.find_first_of("0123456789");
end1 = s.find_last_of("0123456789");
pos2 = ss.find_first_of("0123456789");
end2 = ss.find_last_of("0123456789");
//查找前面的
mid = s.substr(pos1, end1 - pos1 + 1);
int j = pos1 - 2;
for (int i = 0; i < pos1 / 2 - 1; i++) {
map<string, string>::iterator it = arrcnt[s[j] - 'A'][0].find(mid);
if (it != arrcnt[s[j] - 'A'][0].end()) {
if (stoll(it->first) >= arrlen[s[j] - 'A'])
return 0;
mid = it->second;
j -= 2;
continue;
}
else
return 0;
}
//查找后面的
mid1 = ss.substr(pos2, end2 - pos2 + 1);
int jps;
if (pos2==0) {
if (stoll(mid) >= arrlen[s[0] - 'A'])
return 0;
arrcnt[s[0] - 'A'][0].insert(make_pair(mid, mid1));
return 1;
}
jps = pos2 - 2;
for (int i = 0; (jps >= 0)||(i < pos1 / 2); i++) {
map<string, string>::iterator it = arrcnt[ss[jps] - 'A'][0].find(mid1);
if (it != arrcnt[ss[jps] - 'A'][0].end()) {//查找到了
if (stoll(it->first) >= arrlen[ss[jps] - 'A'])
return 0;
mid1 = it->second;
jps -= 2;
continue;
}
else
return 0;
}
arrcnt[s[0] - 'A'][0].insert(make_pair(mid, mid1));
return 1;//没错返回1
}
int main() {
string s;
while (cin >> s && s != ".") {
clear();
int bug = 0, line = 0;
do {
line++;
int eq_pos = s.find('=');
if (eq_pos == -1)
arrlen[s[0] - 'A'] = stoll(s.substr(2, s.size() - 3));
else
if (!bug&&!de_bug(s.substr(0, eq_pos), s.substr(eq_pos + 1, s.size() - eq_pos - 1)))
bug = line;
} while (cin >> s && s != ".");
printf("%d
", bug);
}
return 0;
}
Uva839,天平
#include<cstdio>
#include<iostream>
using namespace std;
bool cret(int& w) {
int wl, dl, wr, dr;
scanf("%d%d%d%d", &wl, &dl, &wr, &dr);
bool a1 = true, a2 = true;
if (wl == 0)
a1=cret(wl);
if (wr == 0)
a2=cret(wr);
w = wl + wr;
return a1&&a2&&wl*dl == wr*dr;
}
int main() {
int kase, w;
scanf("%d", &kase);
while (kase--) {
if (cret(w))
cout << "YES
";
else
cout << "NO
";
if (kase)
cout << "
";
}
return 0;
}
Uva699,下落的树叶
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 10000;
int sum[maxn];
void build(int pos) {
int v;
scanf("%d", &v);
if (v == -1)
return;
sum[pos] += v;
build(pos - 1);
build(pos + 1);
}
bool ini() {
int v;
scanf("%d", &v);
if (v == -1)
return false;
memset(sum, 0, sizeof(sum));
int pos = maxn / 2;
sum[pos] = v;
build(pos - 1);
build(pos + 1);
}
int main() {
int kase = 0;
while (ini()) {
int pos = maxn / 2;
while (sum[pos] != 0)
pos--;
++pos;
printf("Case %d:
", ++kase);
bool i = 0;
while (sum[pos] != 0)
{
if (i != 1)
i = 1;
else
cout << " ";
printf("%d", sum[pos++]);
}
printf("
");
}
return 0;
}
Uva1103,埃及文字(dfs求联通快)
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<cstring>
using namespace std;
char bin[256][5];
const int maxh = 200 + 5;
const int maxw = 50 * 4 + 5;
int H, W, pic[maxh][maxw], color[maxh][maxw];//pic为像素矩阵,color为标记数组
char line[maxw];
const int dr[] = { -1,1,0,0 };
const int dc[] = { 0,0,-1,1 };
vector<set<int>>neighbors;
const char* code = "WAKJSD";
void decode(char ch, int r, int c) {
for (int i = 0; i < 4; i++) {
pic[r][c + i] = bin[ch][i] - '0';
}
}
void dfs(int row, int col, int c) {
color[row][col] = c;
for (int i = 0; i < 4; i++) {
int row2 = row + dr[i];
int col2 = col + dc[i];
if (row2 >= 0 && row2 < H && col2 >= 0 && col2 < W&&pic[row2][col2] == pic[row][col] && color[row2][col2] == 0) {
dfs(row2, col2, c);
}
}
}
//查看一个颜色的块周围的其他颜色的快有多少种,1为白色,不算
void check_neighbors(int row,int col) {
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
int row2 = row + dr[i];
int col2 = col + dc[i];
if (row2 >= 0 && row2 < H && col2 >= 0 && col2 < W && pic[row2][col2] == 0 && color[row2][col2] != 1)
neighbors[color[row][col]].insert(color[row2][col2]);
}
}
}
char recognize(int c) {
int cnt = neighbors[c].size();
return code[cnt];
}
int main(){
strcpy(bin['0'], "0000");
strcpy(bin['1'], "0001");
strcpy(bin['2'], "0010");
strcpy(bin['3'], "0011");
strcpy(bin['4'], "0100");
strcpy(bin['5'], "0101");
strcpy(bin['6'], "0110");
strcpy(bin['7'], "0111");
strcpy(bin['8'], "1000");
strcpy(bin['9'], "1001");
strcpy(bin['a'], "1010");
strcpy(bin['b'], "1011");
strcpy(bin['c'], "1100");
strcpy(bin['d'], "1101");
strcpy(bin['e'], "1110");
strcpy(bin['f'], "1111");
//边读入边解码
int kase = 0;
while (scanf("%d%d", &H, &W) == 2 && H) {
memset(pic, 0, sizeof(pic));
for (int i = 0; i < H; i++) {
scanf("%s", line);
for (int j = 0; j < W; j++) {
decode(line[j], i + 1, j * 4 + 1);
}
}
H += 2;
W = W * 4 + 2;
//求联通快,并依次标记编号
int cnt = 0;
vector<int>cc;
memset(color, 0, sizeof(color));
for (int i = 0; i < H; i++) {
for (int j = 0; j < W; j++) {
if (!color[i][j]) {
dfs(i, j, ++cnt);
if (pic[i][j] == 1)
cc.push_back(cnt);
}
}
}
neighbors.clear();
neighbors.resize(cnt + 1);
for (int i = 0; i < H; i++) {
for (int j = 0; j < W; j++) {
if (pic[i][j] == 1)
check_neighbors(i, j);
}
}
vector<char>ans;
for (int i = 0; i < cc.size(); i++) {
ans.push_back(recognize(cc[i]));
}
sort(ans.begin(), ans.end());
printf("Case %d: ", ++kase);
for (int i = 0; i < ans.size(); i++) {
printf("%c", ans[i]);
}
printf("
");
}
return 0;
}
Uva572,油田(dfs求联通快)
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 100 + 5;
char pic[maxn][maxn];
int m, n, id[maxn][maxn];
void dfs(int row, int col, int c) {
if (row < 0 || row>=m || col < 0 || col>=n) {
return;
}
if (id[row][col] > 0 || pic[row][col] != '@') {
return;
}
id[row][col] = c;
for (int i = -1; i<2; i++) {
for (int j = -1; j<2; j++) {
if (i != 0 || j != 0) {
dfs(row+i, col+j, c);
}
}
}
}
int main() {
while (scanf("%d %d", &m, &n) == 2 && n&&m) {
for (int i = 0; i<m; i++) {
scanf("%s", pic[i]);
}
memset(id, 0, sizeof(id));
int cnt = 0;
for (int i = 0; i<m; i++) {
for (int j = 0; j<n; j++){
if (id[i][j] == 0 && pic[i][j] == '@')
dfs(i, j, ++cnt);
}
}
printf("%d
", cnt);
}
return 0;
}
Uva816,Abott的复仇(dfs加抽象路径判断)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
FILE* fp = fopen("stdout.txt", "w");
struct Node {
int r, c, dir;
Node(int r = 0, int c = 0, int dir = 0) :r(r), c(c), dir(dir) {}
};
const int maxn = 10;
const char* dirs = "NESW";
const char* turns = "FLR";
int eage[maxn][maxn][4][3];
int d[maxn][maxn][4];
Node p[maxn][maxn][4];
int r0, c0, dir, r1, c1, r2, c2;
int dir_id(char c) { return strchr(dirs, c) - dirs; }
int turn_id(char c) { return strchr(turns, c) - turns; }
const int dr[] = { -1,0,1,0 };
const int dc[] = { 0,1,0,-1 };
Node walk(const Node&u, int turn) {
int dir = u.dir;
if (turn == 1)dir = (dir + 3) % 4;
if (turn == 2)dir = (dir + 1) % 4;
return Node(u.r + dr[dir], u.c + dc[dir], dir);
}
bool read() {
char s[99], ss[99];
if (scanf("%s%d%d%s%d%d", s, &r0, &c0, ss, &r2, &c2) != 6)
return false;
//printf("%s
", s);
fprintf(fp,"%s
", s);
dir = dir_id(ss[0]);
r1 = r0 + dr[dir];
c1 = c0 + dc[dir];
memset(eage, 0, sizeof(eage));
for (;;) {
int r, c;
scanf("%d", &r);
if (r == 0)
break;
scanf("%d", &c);
while (scanf("%s", s) == 1 && s[0] != '*')
for (int i = 1; i < strlen(s); i++)
eage[r][c][dir_id(s[0])][turn_id(s[i])] = 1;
}
return true;
}
bool inside(int r,int c) {
return r >= 1 && r <= 9 && c >= 1 && c <= 9;
}
void print(Node uu) {
vector<Node>nodes;
for (;;) {
nodes.push_back(uu);
if (d[uu.r][uu.c][uu.dir] == 0)
break;
uu = p[uu.r][uu.c][uu.dir];
}
nodes.push_back(Node(r0, c0, dir));
int cnt = 0;
for (int i = nodes.size() - 1; i >= 0; i--) {
if (cnt % 10 == 0)
//printf(" ");
fprintf(fp, " ");
//printf(" (%d,%d)", nodes[i].r, nodes[i].c);
fprintf(fp, " (%d,%d)", nodes[i].r, nodes[i].c);
if (++cnt % 10 == 0)
//printf("
");
fprintf(fp, "
");
}
if (nodes.size() % 10 != 0)
//printf("
");
fprintf(fp, "
");
}
void solve() {
queue<Node>q;
memset(d, -1, sizeof(d));
Node u(r1, c1, dir);
d[u.r][u.c][u.dir] = 0;
q.push(u);
while (!q.empty()) {
Node uu = q.front(); q.pop();
if (uu.r == r2&&uu.c == c2) {
print(uu);
return;
}
for (int i = 0; i < 3; i++) {
Node v = walk(uu, i);
if (eage[uu.r][uu.c][uu.dir][i] && inside(v.r, v.c) && d[v.r][v.c][v.dir] < 0) {
d[v.r][v.c][v.dir] = d[uu.r][uu.c][uu.dir] + 1;
p[v.r][v.c][v.dir] = uu;
q.push(v);
}
}
}
//printf(" No Solution Possible
");
fprintf(fp, " No Solution Possible
");
}
int main() {
while (read()) {
solve();
}
fclose(fp);
return 0;
}