You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.
The second line contains n distinct digits a1, a2, ..., an (1 ≤ ai ≤ 9) — the elements of the first list.
The third line contains m distinct digits b1, b2, ..., bm (1 ≤ bi ≤ 9) — the elements of the second list.
Print the smallest pretty integer.
2 3 4 2 5 7 6
25
8 8 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
1
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.
题意就是输出一个最小数,这个数最少要包含一个第一序列的数,和一个第二序列的数。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 0;
inline int get(){
char c;
while((c = getchar()) < '0' || c > '9');
int cnt = c - '0';
while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
return cnt;
}
bool a[20],b[20];
int mina,minb;
int n,m,ans;
int main(){
/* #ifdef lwy
#else
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif*/
memset(a,false,sizeof(a));
memset(b,false,sizeof(b));
n = get(); m = get();
mina = 10; minb = 10;
for(int i = 1; i <= n; i++){
int x; x = get();
mina = min(mina,x);
a[x] = true;
}
for(int i = 1; i <= m; i++){
int x; x = get();
minb = min(minb,x);
b[x] = true;
}
for(int i = 1; i <= 9; i++){
if(a[i] && b[i]){
ans = i;
printf("%d",ans);
return 0;
}
}
ans = min(mina,minb) * 10 + max(mina,minb);
printf("%d",ans);
return 0;
}
题意是将一个序列分成k段,要求每段的最小值最大,并输出这个最大值。
一开始以为是个简单的二分。
但再仔细一看其实只要把k分成 == 1 ,== 2 ,>= 3,3类分类讨论就行了。
然而一开始没有考虑到 == 2 与 >=3 的区别,被hacked了。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1E5;
inline int get(){
char c;
while((c = getchar()) < '0' || c > '9');
int cnt = c - '0';
while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
return cnt;
}
int N,K;
int a[MAXN+10];
int b[MAXN+10],c[MAXN+10];
int maxa,mina,ans;
int main(){
/* #ifdef lwy
#else
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif*/
N = get(); K = get();
for(int i = 1; i <= N; i++){
scanf("%d",&a[i]);
}
mina = maxa = a[1];
for(int i = 1; i <= N; i++){
mina = min(mina,a[i]);
maxa = max(maxa,a[i]);
}
if(K == 1){
ans = mina;
}
if(K >= 3){
ans = maxa;
}
if(K == 2){
ans = mina;
for(int i = 1; i <= N; i++){
b[i] = c[i] = a[i];
}
for(int i = 2; i <= N; i++){
b[i] = min(b[i],b[i-1]);
}
for(int i = N-1; i >= 1; i--){
c[i] = min(c[i],c[i+1]);
}
for(int i = 1; i <= N-1; i++){
ans = max(max(b[i],c[i+1]),ans);
}
}
printf("%d",ans);
return 0;
}
You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.
q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
1 12
3
2 6 8
1 2
3 1 2 3
-1 -1 -1
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings.
题意就是要将一个数分成几个合数之和,求最大的合数个数。
这题显然一个数分成4k+n时会最多
把这个数取4的模
分4类分别讨论就行了。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 0;
inline int get(){
char c;
while((c = getchar()) < '0' || c > '9');
int cnt = c - '0';
while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
return cnt;
}
int q,n;
int k,t,ans;
int main(){
/* #ifdef lwy
#else
freopen(".in","r",stdin);
freopen(".out","w",stdout);
#endif*/
q = get();
while(q--){
n = get();
k = n % 4;
t = n / 4;
if(k == 0) ans = t;
if(k == 1) {
t -= 2;
if(t < 0) ans = -1;
if(t == 0) ans = 1;
if(t > 0) ans = t + 1;
}
if(k == 2){
t -= 1;
if(t < 0) ans = -1;
if(t == 0) ans = 1;
if(t > 0) ans = t + 1;
}
if(k == 3){
t -= 3;
if(t < 0) ans = -1;
if(t == 0) ans = 2;
if(t > 0) ans = t + 2;
}
printf("%d\n",ans);
}
return 0;
}