[CF707 Div2, A ~ D]
（相信进这个博客的人，都已经看过题目了，不再赘述）
这把打小号打到了\(484\),\(rating + 636\)
\(A\)
考虑进行模拟就行了，说白了这是一个英语阅读题
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll T,a[200],b[200],t[200],now;
int main(){
scanf("%lld",&T);
while(T -- ){
ll n;
scanf("%lld",&n);
for(int i = 1;i <= n;++i){
scanf("%lld%lld",&a[i],&b[i]);
}
for(int i = 1;i <= n;++i)
scanf("%lld",&t[i]);
ll now = t[1] + a[1];
for(int i = 1;i <= n - 1;++i){
now += std::ceil(((double)b[i] - a[i]) / 2.0);
now = std::max(now,b[i]);
now += (a[i + 1] - b[i] + t[i + 1]);
}
std::cout<<now<<std::endl;
}
}
\(B\)
考虑进行一个差分处理，再还原原数组
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll t,cnt[200010],now;
int main(){
scanf("%lld",&t);
while(t -- ){
ll n;
scanf("%lld",&n);
for(int i = 0;i <= n + 1;++i)
cnt[i] = 0;
for(int i = 1;i <= n;++i){
ll x = read();
cnt[i + 1] -= 1;
cnt[std::max((ll)0,i - x + 1)] += 1;
}
for(int i = 1;i <= n;++i){
cnt[i] = cnt[i - 1] + cnt[i];
if(cnt[i])
std::cout<<1<<" ";
else
std::cout<<0<<" ";
}
puts("");
}
}
\(C\)
大毒瘤题，想了\(1h\)的做法最后和暴力一样写
观察\(n\)的范围和值域，我们发现在\(n\)较大时，点对数\((n * (n - 1)) / 2 > 2 * |V|\),所以必定有解，那么我们利用这个来进行一个抽屉原理证明
可以证明暴力写法复杂度是对的
// code by Dix_
#include<bits/stdc++.h>
#define ll long long
inline ll read(){
char C=getchar();
ll N=0 , F=1;
while(('0' > C || C > '9') && (C != '-')) C=getchar();
if(C == '-') F=-1 , C=getchar();
while('0' <= C && C <= '9') N=(N << 1)+(N << 3)+(C - 48) , C=getchar();
return F*N;
}
ll n;
std::map<ll,int>QWQ;
std::map<ll,int>QAQ;
ll num[200005];
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i)
scanf("%lld",&num[i]);
for(int i = 1;i <= n;++i)
for(int j = i + 1;j <= n;++j){
ll sum = num[i] + num[j];
if(QWQ.count(sum))
if(QWQ[sum] != i && QAQ[sum] != j && QWQ[sum] != j && QAQ[sum] != i){
puts("YES");
std::cout<<QWQ[sum]<<" "<<QAQ[sum]<<" "<<i<<" "<<j<<std::endl;
return 0;
}
QWQ[sum] = i,QAQ[sum] = j;
}
puts("NO");
}
\(D\)
考虑一个周期内有多少个不同的，再二分求出后面零散的天数
\(Excrt\)不会，得去学，这题是后来补的
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=1e6+5;
LL muler(LL x,LL k,LL MOD)
{
LL res=0; x=(x%MOD+MOD)%MOD; k=(k%MOD+MOD)%MOD;
while(k) {
if(k&1) res=(res+x)%MOD;
x=(x+x)%MOD; k>>=1;
}
return res%MOD;
}
LL power(LL x,LL k,LL MOD)
{
LL res=1; x%=MOD;
while(k) {
if(k&1) res=muler(res,x,MOD);
x=muler(x,x,MOD); k>>=1;
}
return res%MOD;
}
LL exgcd(LL a,LL b,LL& x,LL& y)
{
if(b==0) {
x=1; y=0;
return a;
}
LL z=exgcd(b,a%b,y,x);
y-=a/b*x;
return z;
}
LL inv(LL x,LL p)
{
LL y,z; exgcd(x,p,y,z);
return (y%p+p)%p;
}
LL excrt(int n,LL b[],LL a[])
{
LL m=a[1],ans=b[1];
for(int i=2;i<=n;i++) {
LL y,z,d=exgcd(m,a[i],y,z);
if((b[i]-ans)%d!=0) return -1;
y=muler(y,(b[i]-ans)/d,a[i]/d);
ans+=y*m;
m=a[i]/d*m;
ans=(ans%m+m)%m;
}
return ans;
}
LL gcd(LL a,LL b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
LL lcm(LL a,LL b) { return a/gcd(a,b)*b; }
int n,m;
LL kth;
int a[N],b[N],c[N];
LL A[N],B[N];
vector<LL> v;
LL calc(LL day)
{
// cerr<<day<<" "<<day-(upper_bound(v.begin(),v.end(),day)-v.begin())<<endl;
if(day<=0) return 0;
else return day-(upper_bound(v.begin(),v.end(),day)-v.begin());
}
int main()
{
// freopen("1.in","r",stdin);
int i,j;
LL x;
scanf("%d%d%lld",&n,&m,&kth);
for(i=0;i<n;i++) scanf("%d",&a[i]);
for(i=0;i<m;i++) scanf("%d",&b[i]);
memset(c,-1,sizeof c);
for(i=0;i<n;i++) c[a[i]]=i;
for(i=0;i<m;i++) {
if(~c[b[i]]) {
j=c[b[i]];
A[1]=n; A[2]=m;
B[1]=j; B[2]=i;
x=excrt(2,B,A);
if(x==-1) continue;
v.push_back(x+1);
// cerr<<x<<endl;
}
}
sort(v.begin(),v.end());
LL times=lcm(n,m)-v.size();
LL turns=(kth-1)/times;
kth-=turns*times;
LL L=-1,R=lcm(n,m),mid;
while(L+1<R) {
mid=(L+R)>>1;
if(calc(mid)>=kth) R=mid;
else L=mid;
}
printf("%lld\n",turns*lcm(n,m)+R);
return 0;
}
