A Anagram
链接：https://www.nowcoder.com/acm/contest/123/A
来源：牛客网
空间限制：C/C++ 65536K，其他语言131072K
64bit IO Format: %lld
题目描述
For example, she can transform “ELLY” to “KRIS” character by character by shifting ‘E’ to ‘K’ (6 operations), ‘L’ to ‘R’ (again 6 operations), the second ‘L’ to ‘I’ (23 operations, going from ‘Z’ to ‘A’ on the 15-th operation), and finally ‘Y’ to ‘S’ (20 operations, again cyclically going from ‘Z’ to ‘A’ on the 2-nd operation). The total number of operations would be 6 + 6 + 23 + 20 = 55. However, to make “ELLY” an anagram of “KRIS” it would be better to change it to “IRSK” with only 29 operations. You are given the strings A and B. Find the minimal number of operations needed to transform A into some other string X, such that X is an anagram of B.
输入描述:
There will be multiple test cases. For each testcase: There is two strings A and B in one line.∣A∣=∣B∣≤50. A and B will contain only uppercase letters from the English alphabet (‘A’-‘Z’).
输出描述:
For each test case, output the minimal number of operations.
输入
ABCA BACA ELLY KRIS AAAA ZZZZ
输出
0 29 100
当时想了好久，想明白就好了，就是贪心。
代码：
#include<bits/stdc++.h>
using namespace std;
int main() {
char a[55], b[55];
while(~scanf("%s%s", a, b)) {
int len = strlen(a), vis[55];
long long sum = 0;
memset(vis, 0, sizeof(vis));
for(int i = 0; i < len; i++) {
int minn = 30, num;
for(int j = 0; j < len; j++) {
if(vis[j]) continue;
int temp;
if(b[j] >= a[i]) temp = b[j] - a[i];
else
temp = (26 - (a[i] - b[j]));
if(temp < minn) {
minn = temp;
num = j;
}
}
sum += minn;
vis[num] = 1;
}
printf("%lld\n", sum);
}
return 0;
}
E Sequence
链接：https://www.nowcoder.com/acm/contest/123/E
来源：牛客网
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
We define an element in a sequence "good", if and only if there exists
(1≤j<i) such that
.
Given a permutation p of integers from 1 to n. Remove an element from the permutation such that the number of "good" elements is maximized.
输入描述:
The input consists of several test cases. The first line of the input gives the number of test cases,. For each test case, the first line contains an integer , representing the length of the given permutation. The second line contains n integersp1,p2,…,pn , representing the given permutation p. It's guaranteed that .
输出描述:
For each test case, output one integer in a single line, representing the element that should be deleted. If there are several answers, output the minimal one.
输入
2 1 1 5 5 1 2 3 4
输出
1 5
思路：首先要明白，一个不好数，它总是它之前(包括它)出现的所有数中最小的；
考虑删除一个好数还是不好数：
删除一个好数，则总的好数数量将减少一个(因为删除它后能影响到的好数仅有它自己)；
删除一个不好数，考虑删除这个不好数后能影响到的好数有几个，那么总的好数数量就减少几个；(被它所影响到的好数(假设目前考虑的好数是a[i])是那些满足 最小{a[0~1-i]}<a[i]<=次小{a[0~n-1]} 的数)（转自：点击打开链接）
代码:
#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
#define inf 0x3f3f3f3f
using namespace std;
const int M =1000010;
ll a[M],first[M],second[M];
bool flag[M];
ll cnt[M];
int main()
{
int T;
scanf("%d",&T);
while(T--){
ll n;
scanf("%lld",&n);
ll now_min=inf;
ll pre_min=inf;
for(ll i=1;i<=n;i++){
scanf("%lld",&a[i]);
first[i]=now_min;
second[i]=pre_min;
if(a[i]<=now_min)
{
pre_min=now_min; now_min=a[i];
}
else
{
if(a[i]<pre_min) pre_min=a[i];
}
}
for(ll i=1;i<=n;i++)
{
if(a[i]<=first[i]) flag[i]=1;
else flag[i]=0;
}
ll k;
for(ll i=1;i<=n;i++)
{
if(flag[i]==1)
{
k=i;
cnt[k]=0;
}
else
{
if(a[i]<=second[i]) cnt[k]++;
}
}
ll ans=inf;
for(ll i=1;i<=n;i++)
{
if(flag[i]==1&&cnt[i]==0) ans=min(ans,a[i]);
}
if(ans==inf)
{
for(ll i=1;i<=n;i++)
{
if(flag[i]==0) ans=min(ans,a[i]);
else if(cnt[i]==1) ans=min(ans,a[i]);
}
}
printf("%lld\n",ans);
}
return 0;
}
链接：https://www.nowcoder.com/acm/contest/123/F
来源：牛客网
空间限制：C/C++ 65536K，其他语言131072K
64bit IO Format: %lld
题目描述
输入描述:
The input consists of several test cases. The first line gives the number of test cases,T(1≤T≤106). For each test case, the input contains one line with 8 integers,.
输出描述:
For each test case, output one line containing one integer, representing the answer.
输入
1 1 1 2 2 3 3 4 4
输出
1
思路：
容次原理，一辈子不会忘记的题....
代码：
#include <bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
ll Query(ll l1,ll r1,ll l2,ll r2)
{
ll l=max(l1,l2),r=min(r1,r2);
return r-l+1>0?r-l+1:0;
}
ll Query2(ll l1,ll r1,ll l2,ll r2,ll l3,ll r3)
{
ll l=max(l1,max(l2,l3)),r=min(r1,min(r2,r3));
return r-l+1>0?r-l+1:0;
}
ll Query3(ll l1,ll r1,ll l2,ll r2,ll l3,ll r3,ll l4,ll r4)
{
ll l=max(max(l1,l2),max(l3,l4)),r=min(min(r1,r2),min(r3,r4));
return r-l+1>0?r-l+1:0;
}
int main()
{
int t;
ll l1,r1,l2,r2,l3,r3,l4,r4;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld%lld%lld%lld%lld%lld%lld",&l1,&r1,&l2,&r2,&l3,&r3,&l4,&r4);
ll sum=1;
sum=sum*(r1-l1+1)%mod;
sum=sum*(r2-l2+1)%mod;
sum=sum*(r3-l3+1)%mod;
sum=sum*(r4-l4+1)%mod;
ll ant=0;
ant=(ant+Query(l1,r1,l2,r2)%mod*(r3-l3+1)%mod*(r4-l4+1)%mod)%mod;
ant=(ant+Query(l2,r2,l3,r3)%mod*(r1-l1+1)%mod*(r4-l4+1)%mod)%mod;
ant=(ant+Query(l3,r3,l4,r4)%mod*(r1-l1+1)%mod*(r2-l2+1)%mod)%mod;
ant=(ant+Query(l4,r4,l1,r1)%mod*(r3-l3+1)%mod*(r2-l2+1)%mod)%mod;
ant=(ant-Query2(l1,r1,l2,r2,l3,r3)*(r4-l4+1)%mod+mod)%mod;
ant=(ant-Query(l1,r1,l2,r2)*Query(l3,r3,l4,r4)%mod+mod)%mod;
ant=(ant-Query2(l1,r1,l2,r2,l4,r4)*(r3-l3+1)%mod+mod)%mod;
ant=(ant-Query2(l2,r2,l3,r3,l4,r4)*(r1-l1+1)%mod+mod)%mod;
ant=(ant-Query(l2,r2,l3,r3)*Query(l1,r1,l4,r4)%mod+mod)%mod;
ant=(ant-Query2(l1,r1,l3,r3,l4,r4)*(r2-l2+1)%mod+mod)%mod;
ant=(ant+3*Query3(l1,r1,l2,r2,l3,r3,l4,r4)%mod+mod)%mod;
printf("%lld\n",(sum-ant+mod)%mod);
}
return 0;
}