第一道：POJ 1608
題意：給出P序列--每個數字表示一個右括號左邊有幾個左括號。要求輸出W序列--每個右括號往左遇到幾個左括號才能找到和其相匹配的左括號。
代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
int r,l,ind;
int a[500];
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,x; scanf("%d",&n);
r = l = 0;
ind = 0;
for (int i = 0; i < n; i++)
{
scanf("%d",&x);
for (int j = r; j < x; j++)
{
a[ind++] = 1;//left
}
r = x;
a[ind++] = 0;
}
int num = 0,tmp;
for (int i = 0; i < ind; i++)
{
tmp = 1;
if(a[i] == 0){
num ++;
for (int j = i-1; j >= 0; j--)
{
if(a[j] == 1){
a[j] = -1;
if(num < n)printf("%d ",tmp);
else printf("%d\n",tmp);
break;
}
else if(a[j] == 0) tmp++;
}
}
}
}
return 0;
}第二道：POJ 2632
題意：模擬機器人的運動，判斷是否有衝突
代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
int dir[5][2] = {{0,0},{0,1},{-1,0},{0,-1},{1,0}};
struct robot{
int x,y;
int face;
}r[105];//1 N ,3 S, 2 W, 4 E
int map[105][105];
int main(){
int t,a,b,n,m;
scanf("%d",&t);
while(t--){
memset(map,0,sizeof(map));
scanf("%d%d",&a,&b);
scanf("%d%d",&n,&m);
char tmp;
for (int i = 1; i <= n; i++)
{
scanf("%d %d %c",&r[i].x,&r[i].y,&tmp);
map[r[i].x][r[i].y] = i;
if(tmp == 'N')r[i].face = 1;
else if(tmp == 'S')r[i].face = 3;
else if(tmp == 'W')r[i].face = 2;
else if(tmp == 'E')r[i].face = 4;
}
int ind,rep,flag = 0;
for (int i = 1; i <= m; i++)
{
scanf("%d %c %d",&ind,&tmp,&rep);
if(flag){continue;}
if(tmp == 'L'){
for (int k = 0; k < rep; k++)
{
if(r[ind].face == 4)r[ind].face = 1;
else r[ind].face ++;
}
}else if(tmp == 'R'){
for (int k = 0; k < rep; k++)
{
if(r[ind].face == 1)r[ind].face = 4;
else r[ind].face --;
}
}else if(tmp == 'F'){
map[r[ind].x][r[ind].y] = 0;
for (int k = 0; k < rep; k++)
{
r[ind].x += dir[r[ind].face][0];
r[ind].y += dir[r[ind].face][1];
if(r[ind].x < 1 || r[ind].x > a){flag = 1;break;}
if(r[ind].y < 1 || r[ind].y > b){flag = 1;break;}
if(map[r[ind].x][r[ind].y]!=0){
flag = 2;
printf("Robot %d crashes into robot %d\n",ind,map[r[ind].x][r[ind].y]);
break;
}
}
if(flag == 1)printf("Robot %d crashes into the wall\n",ind);
if(flag == 0)map[r[ind].x][r[ind].y] = ind;
}
}
if(flag == 0)printf("OK\n");
}
return 0;
}第三道：POJ 1573
題意：也是模擬機器人的運動，判斷是安全走出 還是在途中循環了，循環的要輸出 每次循環幾步。
代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
int dir[5][2] = {{0,0},{-1,0},{0,-1},{1,0},{0,1}};
char map[15][15];
int a,b;
int dirc(char tmp){
if(tmp == 'N') return 1;
else if(tmp == 'S')return 3;
else if(tmp == 'W')return 2;
else if(tmp == 'E')return 4;
return 0;
}
bool out(int x,int y){
if(x < 1 || x > a)return 1;
if(y < 1 || y > b)return 1;
return 0;
}
int vis[15][15];
int main(){
int x,y,st;
while(scanf("%d %d %d%*c",&a,&b,&st)!=EOF,a|b|st){
memset(map,'0',sizeof(map));
for (int i = 1; i <= a; i++) scanf("%s",map[i]+1);
memset(vis,0,sizeof(vis));
x = 1,y = st;
int flag = 1,step = 0,d;
while(flag == 1){
if(vis[x][y] == 0){
vis[x][y] = 1;
d = dirc(map[x][y]);
x += dir[d][0]; y += dir[d][1];
step++;
if(out(x,y)){
flag = 0;
printf("%d step(s) to exit\n",step);
break;
}
}else {//loop
while(1){
if(vis[x][y] < 2){//再從開始loop的地方跑一遍，吧vis加到2
vis[x][y]++;
d = dirc(map[x][y]);
x += dir[d][0]; y += dir[d][1];
}else break;
}
int newstep = 0;
int nx = 1,ny = st;
while(1){//就可以再從最初的起點跑，跑到vis爲2的地方，就知道了真相o(*￣▽￣*)ゞ
if(vis[nx][ny] == 2){
printf("%d step(s) before a loop of %d step(s)\n",newstep,step-newstep);
break;
}
d = dirc(map[nx][ny]);
nx += dir[d][0]; ny += dir[d][1];
newstep++;
}
flag = 0;
}
}
}
return 0;
}第四道：POJ 2993
題意：給你國際象棋裏黑白棋子的位置，輸出棋盤。
代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<ctype.h>
using namespace std;
#define INF 0x3f3f3f3f
int q[9] = {1,1,3,5,7,9,11,13,15};
int p[9] = {1,2,6,10,14,18,22,26,30};
char map[35][35] = {
"+---+---+---+---+---+---+---+---+",
"|...|:::|...|:::|...|:::|...|:::|",
"+---+---+---+---+---+---+---+---+",
"|:::|...|:::|...|:::|.:.|:::|...|",
"+---+---+---+---+---+---+---+---+",
"|...|:::|...|:::|...|:::|...|:::|",
"+---+---+---+---+---+---+---+---+",
"|:::|...|:::|...|:::|...|:::|...|",
"+---+---+---+---+---+---+---+---+",
"|...|:::|...|:::|...|:::|...|:::|",
"+---+---+---+---+---+---+---+---+",
"|:::|...|:::|...|:::|...|:::|...|",
"+---+---+---+---+---+---+---+---+",
"|...|:::|...|:::|...|:::|...|:::|",
"+---+---+---+---+---+---+---+---+",
"|:::|...|:::|...|:::|...|:::|...|",
"+---+---+---+---+---+---+---+---+",
};
int main(){
char w[200]={},b[200]={};
gets(w);
int length = strlen(w),x,y;
for (int i = 7; i < length; i++)
{
if(isupper(w[i]) && (w[i-1]==' '||w[i-1]==',')){
x = w[++i]-'a'+1;
y = '8'-w[++i]+1;
map[q[y]][p[x]] = w[i-2];
}else if(islower(w[i]) && w[i-1] == ','){
x = w[i]-'a'+1;
y = '8'-w[++i]+1;
map[q[y]][p[x]] = 'P';
}
}
gets(b);
for (int i = 7; i < length; i++)
{
if(isupper(b[i]) && (b[i-1]==' '||b[i-1]==',')){
x = b[++i]-'a'+1;
y = '8'-b[++i]+1;
map[q[y]][p[x]] = b[i-2]+32;
}else if(islower(b[i]) && b[i-1] == ','){
x = b[i]-'a'+1;
y = '8'-b[++i]+1;
map[q[y]][p[x]] = 'p';
}
}
for (int i = 0; i < 17; i++)
{
printf("%s\n",map[i]);
}
//system("pause");
return 0;
}第五題：POJ 2996
題意：和上題反過來，給出棋盤，輸出棋子位置，注意有輸出格式要求。
代碼：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<ctype.h>
using namespace std;
#define INF 0x3f3f3f3f
const int length = 33;
char col[9] = {0,'a','b','c','d','e','f','g','h'};
int row[9] = {0,1,2,3,4,5,6,7,8};
struct player{
int k[2],q[2],r[2][2],b[2][2],n[2][2],p[8][2];
}W,B;
int main(){
int num = 8,i,j;
char tmp[35],a[35];
int k[2]={0,0},q[2]={0,0},r[2]={0,0},b[2]={0,0},n[2]={0,0},p[2]={0,0};
while(num > 0){
gets(tmp);
gets(a);
for (i = 2,j = 1; i < length; i+= 4,j++)
{
if(islower(a[i])){
if(a[i] == 'k'){
B.k[0] = num; B.k[1] = j;
k[0]++;
}else if(a[i] == 'q'){
B.q[0] = num; B.q[1] = j;
q[0]++;
}else if(a[i] == 'r'){
B.r[r[0]][0] = num; B.r[r[0]][1] = j;
r[0]++;
}else if(a[i] == 'b'){
B.b[b[0]][0] = num; B.b[b[0]][1] = j;
b[0]++;
}else if(a[i] == 'n'){
B.n[n[0]][0] = num; B.n[n[0]][1] = j;
n[0]++;
}else if(a[i] == 'p'){
B.p[p[0]][0] = num; B.p[p[0]][1] = j;
p[0]++;
}
}else if(isupper(a[i])){
if(a[i] == 'K'){
W.k[0] = num; W.k[1] = j;
k[1]++;
}else if(a[i] == 'Q'){
W.q[0] = num; W.q[1] = j;
q[1]++;
}else if(a[i] == 'R'){
W.r[r[1]][0] = num; W.r[r[1]][1] = j;
r[1]++;
}else if(a[i] == 'B'){
W.b[b[1]][0] = num; W.b[b[1]][1] = j;
b[1]++;
}else if(a[i] == 'N'){
W.n[n[1]][0] = num; W.n[n[1]][1] = j;
n[1]++;
}else if(a[i] == 'P'){
W.p[p[1]][0] = num; W.p[p[1]][1] = j;
p[1]++;
}
}
}
num --;
}
gets(tmp);
//print
printf("White: ");
for (i = 0; i < k[1]; i++)printf("K%c%d", col[W.k[1]],row[W.k[0]]);
for (i = 0; i < q[1]; i++)printf(",Q%c%d",col[W.q[1]],row[W.q[0]]);
for (i = 0; i < r[1]; i++)printf(",R%c%d",col[W.r[i][1]],row[W.r[i][0]]);
for (i = 0; i < b[1]; i++)printf(",B%c%d",col[W.b[i][1]],row[W.b[i][0]]);
for (i = 0; i < n[1]; i++)printf(",N%c%d",col[W.n[i][1]],row[W.n[i][0]]);
//輸出格式。
for (i = 0; i < p[1]; i++)
{
for (int k = 0; k < p[1]-i; k++)
{
if(W.p[k][0] > W.p[k+1][0]){
swap(W.p[k][0],W.p[k+1][0]);swap(W.p[k][1],W.p[k+1][1]);
}else if(W.p[k][0] == W.p[k+1][0] && W.p[k][1] > W.p[k+1][1]){
swap(W.p[k][0],W.p[k+1][0]);swap(W.p[k][1],W.p[k+1][1]);
}
}
}
for (i = 0; i < p[1]; i++)printf(",%c%d", col[W.p[i][1]],row[W.p[i][0]]);
printf("\n");
printf("Black: ");
for (i = 0; i < k[0]; i++)printf("K%c%d", col[B.k[1]],row[B.k[0]]);
for (i = 0; i < q[0]; i++)printf(",Q%c%d",col[B.q[1]],row[B.q[0]]);
for (i = 0; i < r[0]; i++)printf(",R%c%d",col[B.r[i][1]],row[B.r[i][0]]);
for (i = 0; i < b[0]; i++)printf(",B%c%d",col[B.b[i][1]],row[B.b[i][0]]);
for (i = 0; i < n[0]; i++)printf(",N%c%d",col[B.n[i][1]],row[B.n[i][0]]);
for (i = 0; i < p[0]; i++)printf(",%c%d", col[B.p[i][1]],row[B.p[i][0]]);
printf("\n");
//system("pause");
return 0;
}╮(╯▽╰)╭，5道簡單的模擬題，本以爲能AK，現實就是打臉，編的慢，慢，慢！！！
