P1211 [USACO1.3]牛式 Prime Cryptarithm
1.3 牛式
题目的意思要读懂,而且啊啊啊。不要轻敌啊啊啊。
昨天的cf也是吃亏在这里,读题读题。
然后需要一个好的思路。
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竟然是因为我没测样例,感觉扔给计算机算就可以了,结果题目意思读错了。
…… 可以n个数字不全用上,还改了半天。
或许写个小函数是个好习惯。
好孩子不要像我这么写。。 一怎么样就太容易慌了
太慌了emmm 中间可以用小函数替代的
#include<iostream>
#include<string>
#include<set>
using namespace std;
int s[15];
bool a[15];
set<int> ss;
//这是好习惯吗 能开bool 就开bool啊
int main() {
int n; cin >> n; int b;
for (int i = 1; i <= n; i++) {
cin >> b;
s[b]++;
ss.insert(b);
}
int cnt = 0;
for (int i = 100; i <= 999; i++) {
for (int j = 10; j <= 99; j++) {
// cout << i << j << endl;
// i = 468;j = 23;
int tmp1 = j / 10; int tmp2 = j % 10;// 1是十位 2 是个位
//cout << tmp1 * i<< " "<<tmp2 * i<<" "<< tmp1 * 10 * i + tmp2 * i<<endl;
if (tmp1*i > 999 || tmp2 * i > 999 || (tmp1 * 10 * i + tmp2 * i) >9999)continue;
else {// 合法 判断是不是那些数字组成的
int t1 = i / 100;
int t2 = (i - t1 * 100) / 10;// 182-100 82/10=8
int t3 = (i - (t1 * 100 + t2 * 10));// 182-100-80=2
if ((ss.count(tmp1) != 1) || (ss.count(tmp2) != 1) || (ss.count(t1) != 1) || (ss.count(t2) != 1) || (ss.count(t3) != 1))continue;
set<int>sss;
sss.insert(t1); sss.insert(t2); sss.insert(t3); sss.insert(tmp1); sss.insert(tmp2);
int k = tmp1 * i;
int k1 = k / 100;
int k2 = (k - k1 * 100) / 10;// 182-100 82/10=8
int k3 = (k - (k1 * 100 + k2 * 10));// 182-100-80=2
if ((ss.count(k1) != 1) || (ss.count(k2) != 1) || (ss.count(k3) != 1) )continue;
sss.insert(k1); sss.insert(k2); sss.insert(k3);
int kk = tmp2 * i;
int kk1 = kk / 100;
int kk2 = (kk - kk1 * 100) / 10;// 182-100 82/10=8
int kk3 = (kk - (kk1 * 100 + kk2 * 10));// 182-100-80=2
if ((ss.count(kk1) != 1) || (ss.count(kk2) != 1) || (ss.count(kk3) != 1))continue;
sss.insert(kk1); sss.insert(kk2); sss.insert(kk3);
int kkk = tmp1 * 10 * i + tmp2 * i;
int kkk1 = kkk / 1000;
int kkk2 = (kkk - kkk1 * 1000) / 100;// 2345- 2000 3
int kkk3 = (kkk - kkk1 * 1000 - kkk2 * 100) / 10;
int kkk4 = (kkk - kkk / 10 * 10);// 2345 - 2340
//cout << kkk << endl;
//cout << kkk1 << kkk2 << kkk3 << kkk4<<endl;
if ((ss.count(kkk1) != 1) || (ss.count(kkk2) != 1) || (ss.count(kkk3) != 1))continue;
sss.insert(kkk1); sss.insert(kkk2); sss.insert(kkk3);
//if (sss.size() != ss.size())continue;
cnt++;
/*int t1 = i /100;
int t2 = (i - t1 * 100) / 10;// 182-100 82/10=8
//..228 -200 28/10
int t3 = (i - (t1 * 100 + t2 * 10));// 182-100-80=2
printf(" %d %d %d %d %d \n" ,tmp1,tmp2,t1,t2,t3);
//************注意一下
//*********** 怎么判断 abcde aabcd
memset(a, 0, sizeof(a));
int sum = 0;
sum += s[t1];
a[t1] = 1;
if (a[t2] == 1);
else {sum += s[t2];a[t2] = 1;}
if (a[t3] == 1);
else {sum += s[t3]; a[t3] = 1; }
if (a[tmp1] == 1);
else { sum += s[tmp1]; a[tmp1] = 1; }
if (a[tmp2] == 1);
else { sum += s[tmp2]; a[tmp2] = 1; }
if (sum == 5)cnt++;
cout << sum << endl;*/
}
}
}
cout << cnt << endl;
return 0;
}
P1201 [USACO1.1]贪婪的送礼者Greedy Gift Givers
1.1.2 的 用map做
需要注意,printf不能那样读入字符串,cin可以,然后ctrl+z是读入文件末尾(星号)
就像是数组一样,然后前面的 是唯一辨识符
#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
//const int maxn 1e5 + 5;
int d[20005];
int main() {
int n; cin >> n;
map<string, int> m ;
string s[15];
for (int i = 1; i <= n; i++) {
cin >> s[i];
m[s[i]] = 0;
}
//名为 s[i]这个人 的钱是这些
string ss[15];
string s1; int all, mm;
while (cin>>s1>>all>>mm){//((scanf("%s %d %d",s1,&all,&mm)) != EOF) {
if ( mm == 0)continue;
int mon = all / mm;
int temp = all - mon * mm;
//cout << temp << endl;//剩下的多了一些一点点钱
m[s1] = m[s1] - all + temp;
for (int i = 1; i <= mm; i++) {
cin >> s1;
m[s1] += mon;
}//cout << "111" << endl;
}
for (int i = 1; i <= n; i++) {
cout << s[i]<<" "<<m[s[i]]<<endl;
}
return 0;
}
P1203 [USACO1.1]坏掉的项链Broken Necklace
哇,数据量并不大,输入的时候直接左边+n 右边+n 用3n的去存 再算往前/往后就好啦
需要注意的是 区间是我们人为扩大的。所以最后如果扩大的超过了限度,比如全是w的字符串,就会读的越界。
解决办法一个是求前面的和的时候i j有范围最多到n,一个是前面后面加起来超过了n就是可以串起来数全,那么直接取n就可以
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
char s[1100];
int n;
int before(int k) {
int cnt = 0; bool flag = false;
char cc;
for (int i = k; i > k-n; i--) {
if (s[i] == 'w')cnt++;
else if (flag == false) {
cc=s[i] ;
flag = true;
cnt++;
}
else ///flag== true
{
if (s[i] == cc)cnt++;
else return cnt;
}
}
return cnt;
}
int after(int k) {
int cnt = 0; bool flag = false;
char cc;
for (int i = k+1; i<k+1+n; i++) {
if (s[i] == 'w')cnt++;
else if (flag == false) {
cc = s[i];
flag = true;
cnt++;
}
else ///flag== true
{
if (s[i] == cc)cnt++;
else return cnt;
}
}
return cnt;
}
int main() {
char c; cin >> n;
for (int i = 1; i <= n; i++) {
cin >> c;
s[i] = c; s[i + n] = c; s[i + n * 2] = c;
}
//cout << before(2*n-1) + after(2 * n-1);
int maxx = 0;
for (int i = n + 1; i <= n * 2; i++) {
int tmp1 = before(i) + after(i);
if (tmp1 >= n)tmp1 = n;
maxx = max(maxx,tmp1);
}
cout << maxx << endl;
return 0;
}
P1205 [USACO1.2]方块转换 Transformations
这题其实没什么难度,zj说它很经典,但是有很多小问题要注意 …… 其实任何一个题都有很多小问题要注意吧…… 这就提示了我,要好好利用草稿纸,尽量把细节的东西写的清楚,才不至于一直犯错(不说了,我去淘宝买草稿纸了) 【关于如何旋转】 其实在草稿纸上写个小正方形,然后 然后!!!画两个箭头,然后分别看看i和j是如何增大的就行啦,不复杂。 中心对称亦然,两边儿相等一下就行啦。 需要注意的是5这个操作,我比的时候因为都是bool() 所以为了比s1和s2,设立了个s3来改变。 问题是no6的不变不能比较了。。 所以改变之后又加了回去。 其他情况自己debug一下不复杂的。 ……
其他没什么坑啊,自己的程序写的不好 都划水了半年了,找==和=找了半小时导致数据都读入成0了,所有情况理所当然对了…… 扶额 半年了还在做这种题-。- 真是羞愧啊-。- (还做了这么久(小声bb))
#include<iostream>
#include<string>
using namespace std;
int n; int s1[15][15]; int s2[15][15];
char c;
bool no1(){
int i2= 1;
int j2 = n;
for (int i = 1; i <= n; i++) {
int i2= 1;
for (int j = 1; j <= n; j++) {
//cout << s1[i][j] << s2[i][j];
if (s1[i][j] != s2[i2++][j2])return false;
}j2--;
}
return true;
}
bool no2() {
int i2=n;
int j2 = n;
for (int i = 1; i <= n; i++) {
int j2 = n;
for (int j = 1; j <= n; j++) {
if (s1[i][j] != s2[i2][j2--])return false;
}i2--;
}
return true;
}
bool no3() {
int i2 = n;
int j2 = 1;
for (int i = 1; i <= n; i++) {
i2 = n;
for (int j = 1; j <= n; j++) {
if (s1[i][j] != s2[i2--][j2])return false;
}j2++;
}
return true;
}
bool no4() {
int i1 = 1; int j1 = n; int len = n / 2;
for (int i = 1; i <= n; i++) {
int j = 1;
for (int ii = n; ii > len; ii--) {
if (s1[i][j] != s2[i][n + 1 - j])return false;
j++;
}
for (int ii = len; ii >= 1; ii--) {
if (s1[i][j] != s2[i][ii])return false;
j++;
}
i1++;
}
return true;
}
bool no5() {
int s3[15][15];
int i1 = 1; int j1 = n; int len = n / 2;
for (int i = 1; i <= n; i++) {
int j = 1;
for (int ii = n; ii > len; ii--) {
s3[i][j] = s1[i][n + 1 - j];
j++;
}
for (int ii = len; ii >= 1; ii--) {
s3[i][j] = s1[i][ii];
j++;
}
i1++;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
s1[i][j] = s3[i][j];
}
}
bool flag = true;
if (no1() || no2() || no3())flag= true;
else flag = false;
i1 = 1; j1 = n; len = n / 2;
for (int i = 1; i <= n; i++) {
int j = 1;
for (int ii = n; ii > len; ii--) {
s1[i][j] = s3[i][n + 1 - j];
j++;
}
for (int ii = len; ii >= 1; ii--) {
s1[i][j] = s3[i][ii];
j++;
}
i1++;
}
return flag;
}
bool no6() {
int i2=1;
int j2 = 1;
for (int i = 1; i <= n; i++) {
j2 = 1;
for (int j = 1; j <= n; j++) {
if (s1[i][j] != s2[i2][j2])return false;
j2++;
}
i2++;
}
return true;
}
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cin >> c;// s[i][j];
//cout << c << endl;
if (c == '@')//cout << "aaa";
s1[i][j] = 1;
else s1[i][j] = 0;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
cin >> c;// s[i][j];
if (c == '@')s2[i][j] = 1;
else s2[i][j] = 0;
}
}
if (no1()== true)cout << "1" << endl;
else if (no2() == true)cout << "2" << endl;
else if (no3() == true)cout << "3" << endl;
else if (no4() == true)cout << "4" << endl;
else if (no5() == true)cout << "5" << endl;
else if (no6() == true)cout << "6" << endl;
else cout << "7" << endl;
return 0;
}