补补紫书
第3章
例题部分
例题3-1 TEX Quotes UVa272
简单的输入输出
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main(){
int flag = false;
char c;
while( (c=getchar()) !=EOF){
if(c == '"'){
if(!flag){
printf("``");
flag = true;
}
else{
printf("''");
flag = false;
}
}else
printf("%c", c);
}
return 0;
}
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擅用数组
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 110;
char save[maxn] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./\0";
int main(){
char c;
int len = sizeof(save);
int i;
while((c=getchar())!=EOF){
for(i=0; i<len&&save[i]!=c; i++);
printf("%c", i==len?c:save[i-1]);
}
return 0;
}
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例题3-3 Palindromes UVa401
回文与镜像操作
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char rev_ch[] = "A 3 HIL JM O 2TUVWXY5";
char rev_in[] = "01SE Z 8 ";
const int maxn = 1010;
char save[maxn];
int main(){
while(~scanf("%s", save)){
int len = strlen(save);
int half = (len-1)/2;
bool pali = true, mirr = true;
for(int i=0; i<=half; i++){
if(save[i]!=save[len-1-i]) pali = false;
if('A'<=save[i]&&save[i]<='Z'){
if(save[len-i-1] != rev_ch[save[i]-'A']) mirr = false;
}else{
if(save[len-i-1] != rev_in[save[i]-'0']) mirr = false;
}
}
if(pali){
if(mirr){
printf("%s -- is a mirrored palindrome.", save);
}else{
printf("%s -- is a regular palindrome.", save);
}
}else{
if(mirr){
printf("%s -- is a mirrored string.", save);
}else{
printf("%s -- is not a palindrome.", save);
}
}
printf("\n\n");
}
return 0;
}
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例题3-4 Master-Mind Hints UVa340
普通计数
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1010;
int save[maxn];
int in[maxn];
int num[20];
int num_cp[20];
int main(){
int n, T=0;
while(~scanf("%d", &n)){
if(n==0) break;
++T;
printf("Game %d:\n", T);
for(int i=0; i<10; i++) num[i] = 0;
for(int i=0; i<n; i++){
scanf("%d", &save[i]);
++num[save[i]];
}
while(1){
memcpy(num_cp, num, sizeof(num));
bool end = true;
int match=0, misMatch=0;
for(int i=0; i<n; i++){
scanf("%d", &in[i]);
if(in[i] != 0) end = false;
if(in[i]==save[i]){
++match;
--num_cp[in[i]];
}
}
for(int i=0; i<n; i++){
if(in[i]!=save[i])
if(num_cp[in[i]]){
++misMatch;
--num_cp[in[i]];
}
}
if(end) break;
printf(" (%d, %d)\n", match, misMatch);
}
}
return 0;
}
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例题3-5 Digit Generator UVa1583
打表
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 100050;
int save[maxn];
int gen(int x){
int ans = x;
while(x){
ans += x%10;
x /= 10;
}
return ans;
}
int main(){
memset(save, 0, sizeof(save));
for(int i=1; i<=100000; i++){
int ge = gen(i);
if(save[ge]==0) save[ge] = i;
}
int N;
while(~scanf("%d", &N)){
int now;
for(int i=0; i<N; i++){
scanf("%d", &now);
printf("%d\n", save[now]);
}
}
return 0;
}
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例题3-6 Circular Sequence UVa1584
环状数组
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 110;
char str[maxn];
bool compare(int s1, int s2, int len){
int p1 = s1, p2 = s2, p = 0;
while(p<len){
++p;
if(str[p1%len] < str[p2%len]) return true;
else if(str[p1%len] > str[p2%len]) return false;
else{
++p1;
++p2;
}
}
return true;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%s", str);
int len = strlen(str);
int p = 0, ans = 0;
while(p<len){
if(compare(p, ans, len)) ans = p;
++p;
}
p = ans;
for(int i=0; i<len; i++){
printf("%c", str[p%len]);
++p;
}
printf("\n");
}
return 0;
}
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习题部分
普通计数
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 100;
char str[maxn];
int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%s", str);
int ans = 0, combo = 0;
int len = strlen(str);
for(int i=0; i<len; i++){
if(str[i]=='O'){
combo++;
ans += combo;
if(str[i+1] != 'O'){
combo = 0;
}
}
}
ans += combo;
printf("%d\n", ans);
}
return 0;
}
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简单字符串数字切割
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 110;
char str[maxn];
double getVal(char tp, int num){
if(tp=='C')
return num*12.01;
else if(tp=='H')
return num*1.008;
else if(tp=='O')
return num*16.00;
else
return num*14.01;
}
int str2int(int s, int e){
int sum = 0, p = s;
while(p<=e){
sum *= 10;
sum += str[p]-'0';
++p;
}
return sum;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
double ans = 0.0f;
int num = 0;
scanf("%s", str);
int p = 0, val = 0, len = strlen(str);
char tp;
bool counting = false;
for(int i=0; i<len; i++){
if('A'<=str[i]&&str[i]<='Z'){
p = i+1;
tp = str[i];
if('0'>str[i+1]||str[i+1]>'9'){
ans += getVal(tp, 1);
}
}else{
if(str[i+1]<'0'||str[i+1]>'9'){
val = str2int(p, i);
ans += getVal(tp, val);
}
}
}
printf("%.3lf\n", ans);
}
return 0;
}
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习题3-3 Digit Counting UVa1225
普通地打了个表
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 10010;
int res[maxn][10];
void update(int num){
int index = num;
while(num){
++res[index][num%10];
num /= 10;
}
}
int main(){
int T;
scanf("%d", &T);
memset(res, 0, sizeof(res));
for(int i=1; i<=10000; i++){
memcpy(res[i], res[i-1], 10*sizeof(int));
update(i);
}
while(T--){
int n;
scanf("%d", &n);
for(int i=0; i<=9; i++){
printf("%d%c", res[n][i], i==9?'\n':' ');
}
}
return 0;
}
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习题3-4 Periodic Strings UVa455
最小周期串
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 100;
char str[maxn];
int main(){
int n;
scanf("%d", &n);
while(n--){
scanf("%s", str);
int len = strlen(str), res = 1;
while(res<=len){
if(len%res) {
++res;
continue;
}
int rep = len/res;
--rep;
bool flag = true;
while(rep){
for(int i=0+res*rep, p=0; p<res; i++,p++){
if(str[p] != str[i]){
flag = false;
break;
}
}
--rep;
if(!flag) break;
}
if(flag) break;
++res;
}
printf("%d\n", res);
if(n!=0) printf("\n");
}
return 0;
}
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习题3-5 Puzzle UVa227
输入输出有点卡人… 学会了使用fgets..以前都在使用危险的gets
我每次switch都忘记写break呢
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| #include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 10;
char puzz[maxn][maxn];
int main(){
int sx, sy;
int rnd = 1;
bool first = true;
while(fgets(puzz[0], maxn, stdin) != NULL){
if(strlen(puzz[0])<5 && puzz[0][0]=='Z'){
break;
}
if(first) first = false;
else printf("\n");
for(int i=1; i<5; i++){
fgets(puzz[i], maxn, stdin);
}
for(int i=0; i<5; i++)
for(int j=0; j<5; j++){
if(puzz[i][j] == ' '){
sx = i;
sy = j;
}
}
char op[110];
bool flag = true;
while(~scanf("%s", op)){
getchar();
int len = strlen(op);
bool bre = false;
for(int i=0; i<len; i++){
switch(op[i]){
case 'A':
if(0<=sx-1){
swap(puzz[sx][sy], puzz[sx-1][sy]);
sx -= 1;
}else flag = false;
break;
case 'B':
if(5>sx+1){
swap(puzz[sx][sy], puzz[sx+1][sy]);
sx += 1;
}else flag = false;
break;
case 'L':
if(0<=sy-1){
swap(puzz[sx][sy], puzz[sx][sy-1]);
sy -= 1;
}else flag = false;
break;
case 'R':
if(5>sy+1){
swap(puzz[sx][sy], puzz[sx][sy+1]);
sy += 1;
}else flag = false;
break;
default:
if(op[i]!='0') flag = false;
else bre = true;
}
}
if(bre) break;
}
printf("Puzzle #%d:\n", rnd++);
if(!flag){
printf("This puzzle has no final configuration.\n");
}else{
for(int i=0; i<5; i++){
for(int j=0; j<5; j++){
printf("%c%c", puzz[i][j], j==4?'\n':' ');
}
}
}
}
return 0;
}
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习题3-6 Crossword Answers UVa232
嗯…无感想
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
const int maxn = 15;
char puzz[maxn][maxn];
bool used[maxn][maxn];
vector<pair<int,int> > strBlock;
void add(int x, int y, int &num){
strBlock.push_back(make_pair(x, y));
}
void print(int i){
if(i+1<10) printf(" %d.", i+1);
else if(10<=i+1&&i+1<100) printf(" %d.", i+1);
else printf("%d.", i+1);
}
int main(){
int r, c;
int rnd = 1;
bool first = true;
while(~scanf("%d", &r)){
if(!r) break;
if(first) first = 0;
else printf("\n");
strBlock.clear();
scanf("%d", &c);
int num = 0;
for(int i=0; i<r; i++){
scanf("%s", puzz[i]);
for(int j=0; j<c; j++){
if(puzz[i][j]!='*'){
if(j-1<0 || i-1<0) {
add(i, j, num);
}else if(0<=j-1&&puzz[i][j-1]=='*'){
add(i, j, num);
}else if(0<=i-1&&puzz[i-1][j]=='*'){
add(i, j, num);
}
}
}
}
int len = strBlock.size();
memset(used, 0, sizeof(used));
printf("puzzle #%d:\nAcross\n", rnd++);
for(int i=0; i<len; i++){
int px = strBlock[i].first;
int py = strBlock[i].second;
if(used[px][py]) continue;
print(i);
while(py<c && puzz[px][py]!='*'){
used[px][py] = true;
printf("%c", puzz[px][py]);
++py;
}
printf("\n");
}
printf("Down\n");
memset(used, 0, sizeof(used));
for(int i=0; i<len; i++){
int px = strBlock[i].first;
int py = strBlock[i].second;
if(used[px][py]) continue;
print(i);
while(px<r && puzz[px][py]!='*'){
used[px][py] = true;
printf("%c", puzz[px][py]);
++px;
}
printf("\n");
}
}
return 0;
}
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习题3-7 DNA Consensus String UVa1368
记个数
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
char str[60][1010];
int main(){
int T;
scanf("%d", &T);
while(T--){
int m, n;
scanf("%d%d", &m, &n);
for(int i=0; i<m; i++){
scanf("%s", str[i]);
}
int ans = 0;
for(int j=0; j<n; j++){
int sum[4];
memset(sum, 0, sizeof(sum));
for(int i=0; i<m; i++){
switch(str[i][j]){
case 'A':
++sum[0];
break;
case 'C':
++sum[1];
break;
case 'G':
++sum[2];
break;
case 'T':
++sum[3];
break;
default:
break;
}
}
if(sum[0]>=sum[1] && sum[0]>=sum[2] && sum[0]>=sum[3]){
printf("A");
ans += sum[1]+sum[2]+sum[3];
}else if(sum[1]>=sum[0] && sum[1]>=sum[2] && sum[1]>=sum[3]){
printf("C");
ans += sum[0]+sum[2]+sum[3];
}else if(sum[2]>=sum[0] && sum[2]>=sum[1] && sum[2]>=sum[3]){
printf("G");
ans += sum[0]+sum[1]+sum[3];
}else{
printf("T");
ans += sum[0]+sum[1]+sum[2];
}
}
printf("\n%d\n", ans);
}
return 0;
}
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习题3-8 Repeating Decimals UVa202
手写大数除法的基础?
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
const int maxn = 3010;
int used[maxn];
int ans[100010];
int main(){
int a, b, c;
while(~scanf("%d%d", &a, &b)){
memset(used, 0, sizeof(used));
c = a/b;
printf("%d/%d = %d.", a, b, c);
int rep = 0, sta = 1;
a -= c*b;
while(a%b){
if(used[a]) {
sta = used[a];
break;
}else used[a] = rep+1;
a *= 10;
c = a/b;
ans[++rep] = c;
a -= c*b;
}
bool div = false;
if(rep<=50){
if(a%b==0){
div = true;
for(int i=1; i<=rep; i++) printf("%d", ans[i]);
printf("(0)\n");
}else{
for(int i=1; i<sta; i++) printf("%d", ans[i]);
printf("(");
for(int i=sta; i<=rep; i++) printf("%d", ans[i]);
printf(")\n");
}
}else{
if(a%b==0){
div = true;
for(int i=1; i<=50; i++) printf("%d", ans[i]);
printf("...\n");
}else{
for(int i=1; i<sta; i++) printf("%d", ans[i]);
printf("(");
for(int i=sta; i<=50; i++) printf("%d", ans[i]);
printf("...)\n");
}
}
printf(" %d = number of digits in repeating cycle\n\n", div?1:rep+1-sta);
}
return 0;
}
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习题3-9 All in All UVa10340
初看以为是最大公共子串, 但其实只要遍历母串时看看子串是否匹配就好惹
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
const int maxn = 100010;
char str[maxn], pat[maxn];
int main(){
while(~scanf("%s%s", pat, str)){
int lens = strlen(str), lenp = strlen(pat);
int p = 0;
for(int i=0; i<lens; i++){
if(p<lenp && str[i] == pat[p]) ++p;
}
if(p==lenp) printf("Yes\n");
else printf("No\n");
}
return 0;
}
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先把六个面排序, 筛选出三个面来, 如果能组成长方体的话这三个面的边长有逻辑关系, 写个判断
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn =1010;
pair<int, int> cov[10];
bool vis[maxn];
bool can(pair<int, int> a, pair<int, int> b, pair<int, int> c){
if(a.first==b.first){
if(a.second == c. first){
if(c.second == b.second) return true;
}else if(a.second == c.second){
if(c.first == b.second) return true;
}
}
return false;
}
int main(){
while(~scanf("%d%d", &cov[0].first, &cov[0].second)){
memset(vis, 0, sizeof(vis));
for(int i=1; i<6; i++){
scanf("%d%d", &cov[i].first, &cov[i].second);
}
for(int i=0; i<6; i++){
if(cov[i].first > cov[i].second) swap(cov[i].first, cov[i].second);
}
sort(cov, cov+6);
bool flag = true;
for(int i=0; i<6; i+=2){
if(cov[i]!=cov[i+1]) flag = false;
}
if(!can(cov[0], cov[2], cov[4])) flag = false;
if(flag) printf("POSSIBLE\n");
else printf("IMPOSSIBLE\n");
}
return 0;
}
|
习题3-11 Kickdown UVa1588
wa了很久…玉山啊..边界条件不考虑清楚就写….
code
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1010;
char s1[maxn], s2[maxn];
int main(){
while(~scanf("%s%s", s1, s2)){
int len1 = strlen(s1), len2 = strlen(s2);
if(len1>len2){
swap(s1, s2);
swap(len1, len2);
}
int ans = len1+len2;
bool flag;
for(int i=0; i<len1; i++){
flag = true;
for(int j=0; i+j<len1; j++){
if( (s1[i+j]-'0')+(s2[j]-'0') >3){
flag = false;
break;
}
}
if(flag){
if(ans > i+len2) {
ans = i+len2;
}
}
}
for(int i=0; i<len2; i++){
flag = true;
for(int j=0; i+j<len2 && j<len1; j++){
if( (s1[j]-'0')+(s2[i+j]-'0')>3 ){
flag = false;
break;
}
}
if(flag){
int now = max(i+len1, len2);
if(ans > now){
ans = now;
}
}
}
printf("%d\n", ans);
}
return 0;
}
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习题3-12 Floating-Point Numbers UVa11809
这题好玩, 题面复习了一遍浮点数在计算机内是如何存储的.
上图中8-bit用来存储Mantissa(尾数), 6-bit用于存储exponent(阶码).
本题默认首位为0, 即表示的是正数. 图中8-bit Mantissa+6-bit expnent最大能表示的浮点数为$0.11111111_2\times 2^{111111_2}=0.998046875 \times 2^{63}=9.205357638345293824e18$, 我们记为$M_2\times 2^{E_2}=A\times 10^B$
现在以$A\times 10^B$的形式输入计算机能表示的最大浮点数, 求M, E的位数. 直接算很麻烦, 由于题目明确说明$0\le M\le 9, 1\le E\le 30$, 所以我们可以直接打表, 把10*30种情况打出来. 还有一个问题, $2^{E_2}$会非常大,直接算会变成inf. 此处使用一个log操作:
$log(M_2\times 2^{E_2})=log(A\times 10^B)$
$logM_2+ E_2\times log2=logA+ B$
code
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| #include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double logt = log10(2);
const double eps = 1e-6;
double base[2] = {0.1, 10};
double save[15][35];
double str2dou(char* str, int len){
bool flag = true;
double now = 1, ans = 0;
for(int i=0; i<len; i++){
if(str[i]=='.'){
flag = false;
now = 0.1;
}else{
if(flag){
ans*=10;
ans += str[i]-'0';
}else{
ans += (str[i]-'0')*now;
now *= base[flag];
}
}
}
return ans;
}
void init(){
double sum1 = 0;
double now1 = 0.5;
for(int m=0; m<=9; m++){
sum1 += now1;
now1 /= 2;
double sum2 = 0;
double now2 = 1;
for(int e=1; e<=30; e++){
sum2 += now2;
now2 *= 2;
save[m][e] = log10(sum1) + sum2*logt;
}
}
}
int main(){
char str[110];
init();
while(~scanf("%s", str)){
int pos_e;
int len = strlen(str);
for(pos_e=0; str[pos_e]!='e'; pos_e++);
str[pos_e] = '\0';
char* A = str;
char* B = str+pos_e+1;
int lenA = strlen(A), lenB = strlen(B);
if(lenA==1 && lenB==1 && A[0]=='0' && B[0]=='0') break;
double a = str2dou(A, lenA), b = str2dou(B, lenB);
bool found = false;
for(int m=0; m<=9; m++){
for(int e=1; e<=30; e++){
if(abs(log10(a)+b - save[m][e])<eps){
found = true;
printf("%d %d\n", m, e);
break;
}
}
if(found) break;
}
}
return 0;
}
|
第4章
例题部分
例题4-1 Ancient Cipher UVa1339
时隔一年的排序
code
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| #include <stdio.h>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 110;
int main(){
char a[maxn], b[maxn];
int staa[26], stab[26];
while(~scanf("%s%s", a, b)){
memset(staa, 0, sizeof(staa));
memset(stab, 0, sizeof(stab));
int len = strlen(a);
for(int i=0; i<len; i++){
staa[a[i]-'A']++;
stab[b[i]-'A']++;
}
sort(staa, staa+26);
sort(stab, stab+26);
bool flag = true;
for(int i=0; i<26; i++){
if(staa[i]!=stab[i]){
flag = false;
break;
}
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
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习题4-2 Hangman Judge UVa489
习题部分