链接:http://codeforces.com/contest/625
problemA:很经典的问题,你有n元钱,有两种牛奶,A:a元一瓶,B:b元一瓶,喝完返还瓶子退回c元,问最大能喝多少瓶牛奶。
分析:贪心先全买A或者B,取最大的即可。
代码:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=1020;
const int MAX=151;
const int MOD1=100000007;
const int MOD2=100000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=9999991;
const ll INF=10000000010;
typedef unsigned long long ull;
int main()
{
ll a,b,c,n,ans1=0,ans2=0;
scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &c);
ans1=n/a;
if (n>=b) {
ans2=(n-b)/(b-c);n-=(n-b)/(b-c)*(b-c);
if (n>=b) { ans2++;n-=b-c; }
ans2+=n/a;
}
printf("%I64d\n", max(ans1,ans2));
return 0;
}
problemB:给两个字符串s和t,问最少修改多少个s中的字符使得s中不再存在串t。
分析:kmp跑一遍,然后贪心每次修改最后那一个字符即可。
代码:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=410;
const int MAX=151;
const int MOD1=100000007;
const int MOD2=100000009;
const double EPS=0.00000001;
typedef long long ll;
const ll INF=5000011;
const ll MOD=1000000007;
typedef unsigned long long ull;
int q[100010],nex[35];
char t[100010],s[35];
void pre_kmp(char s[],int les)
{
int i,k;
nex[0]=-1;k=-1;
for (i=0;i<les;i++) {
while (k>-1&&s[k]!=s[i]) k=nex[k];
k++;nex[i+1]=k;
}
}
void kmp(char s[],char t[],int les,int let)
{
pre_kmp(s,les);
memset(q,0,sizeof(q));
int i,k=0,bo=0;
for (i=0;i<let;i++) {
while (k>0&&s[k]!=t[i]) k=nex[k];
if (s[k]==t[i]) k++;
if (k==les) { q[i]=1;k=nex[k]; }
}
}
int main()
{
int i,lent,lens,mx=-1,ans=0;
scanf("%s", t);
scanf("%s", s);
kmp(s,t,strlen(s),strlen(t));
lent=strlen(t);lens=strlen(s);
for (i=0;i<lent;i++)
if (q[i]&&i-mx>=lens) { ans++;mx=i; }
printf("%d\n", ans);
return 0;
}
problemC:给定n,k。要求构造一个用1~n*n填满n*n的矩阵,使得第k列的和最大。
分析:贪心即可。
代码:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=410;
const int MAX=151;
const int MOD1=100000007;
const int MOD2=100000009;
const double EPS=0.00000001;
typedef long long ll;
const ll INF=10000000010;
const ll MOD=1000000007;
typedef unsigned long long ull;
int main()
{
int i,j,l,r,n,k;
scanf("%d%d", &n, &k);
printf("%d\n", (k*n-n+1+n*n-n+k)*n/2);
l=0;r=(k-1)*n;
for (i=1;i<=n;i++) {
for (j=1;j<k;j++) printf("%d ", ++l);
for (j=1;j<=n-k+1;j++) printf("%d ", ++r);
printf("\n");
}
return 0;
}
problemD:给一个长度<=100000的数n,求x使得x+f(x)=n,例f(12309)=90321,x不能有前缀0。无解输出0。
分析:很容易想到答案只有两种解,一种长度和n一样,另一种长度比n小1,如果确定下来长度的话,那么从数字的两端向中间凑即可,多了进位,少了退位。
代码:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=100010;
const int MAX=151;
const int MOD1=100000007;
const int MOD2=100000009;
const double EPS=0.00000001;
typedef long long ll;
const ll INF=10000000010;
const ll MOD=1000000007;
typedef unsigned long long ull;
char s[N];
int len,a[N],b[N];
int check() {
for (int i=1;i<=len/2;) {
int l=i,r=len-i+1;
if (a[l]==a[r]) i++;
else if (a[l]==a[r]+1||a[l]==a[r]+11) {
a[l]--;a[l+1]+=10;
} else if (a[l]==a[r]+10) {
a[r-1]--;a[r]+=10;
} else return 0;
}
if (len&1) {
if (a[len/2+1]%2==1||a[len/2+1]>18||a[len/2+1]<0) return 0;
b[len/2+1]=a[len/2+1]/2;
}
for (int i=1;i<=len/2;i++) {
if (a[i]>18||a[i]<0) return 0;
b[i]=(a[i]+1)/2;b[len-i+1]=a[i]/2;
}
return b[1]>0;
}
void put() {
for (int i=1;i<=len;i++) printf("%d", b[i]);
printf("\n");
}
int main()
{
scanf("%s", s);
len=strlen(s);
for (int i=1;i<=len;i++) a[i]=s[i-1]-'0';
if (check()) {
put();return 0;
}
if (s[0]=='1'&&len>1) {
for (int i=1;i<len;i++) a[i]=s[i]-'0';
a[1]+=10;len--;
if (check()) put();
else printf("0\n");
} else printf("0\n");
return 0;
}