Questions surface: https://www.cnblogs.com/Juve/articles/11615883.html
Country X Army:
It seems to have O (T * N) Direct greedy practices
In fact, with more than half of a log can be over
First of all positions by ba descending order (after this program, the sort order is optimal sweep)
Then half of the answer
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define re register
using namespace std;
const int MAXN=1e5+5;
inline int read(){
re int x=0;re char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0',ch=getchar();}
return x;
}
int t,n,l,r;
struct node{
int a,b;
inline friend bool operator < (node p,node q){
return p.b-p.a>q.b-q.a;
}
}c[MAXN];
inline bool check(re int x){
for(re int i=1;i<=n;++i){
if(x<c[i].b) return 0;
x-=c[i].a;
}
return 1;
}
signed main(){
t=read();
while(t--){
n=read();
l=0,r=0;
for(re int i=1;i<=n;++i){
c[i].a=read(),c[i].b=read();
l+=c[i].a,r+=c[i].b;
}
sort(c+1,c+n+1);
while(l<r){
re int mid=(l+r)>>1;
if(check(mid)) r=mid;
else l=mid+1;
}
printf("%lld\n",l);
}
return 0;
}
Permutations:
The $ C_ {n} ^ {i} * C_ {n} ^ {i} $ into $ C_ {n} ^ {i} * C_ {n} ^ {ni} $,
In this case, that is, for each i, among n number of programs selected by the i-th of the n selected (ni) of the program number, the last add up.
Such answers obtained, in fact, equivalent to 2n of the article, a first n-i selected, selected after n (ni) th ,,
Also, because i runs over all integers from 0 to n, then the number of the accumulated programs is equal to $ C_ {2 * n} ^ {n} $.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define re register
using namespace std;
const int MAXN=2e6+5;
const int mod=1e9+7;
int t,n,fac[MAXN],inv[MAXN];
inline int q_pow(re int a,re int b,re int p){
re int res=1;
while(b){
if(b&1) res=res*a%p;
a=a*a%p;
b>>=1;
}
return res;
}
inline void get_c(re int N){
fac[0]=fac[1]=inv[0]=1;
for(re int i=2;i<=N;++i){
fac[i]=fac[i-1]*i%mod;
}
inv[N]=q_pow(fac[N],mod-2,mod);
for(re int i=N-1;i>=1;--i){
inv[i]=inv[i+1]*(i+1)%mod;
}
}
inline int C(re int n,re int m){
if(m>n) return 0;
if(m==n) return 1;
return fac[n]%mod*inv[m]%mod*inv[n-m]%mod;
}
signed main(){
get_c(2e6);
scanf("%lld",&t);
while(t--){
scanf("%lld",&n);
printf("%lld\n",C(2*n,n));
}
return 0;
}
Palindrome:
Definition of G [i] [j] indicates whether the i to j palindrome this section, may be formed g [i + 1] [j-1] Transfer:
for(int i=1;i<=len;++i){
g[i][i]=1;
f[i][i]=1;
for(int j=1;j<=min(i-1,len-i);++j){
if(s[i-j]==s[i+j]){
g[i-j][i+j]=1;
}else break;
}
}
for(int i=1;i<=len-1;++i){
if(s[i]==s[i+1]){
g[i][i+1]=1;
f[i][i+1]=3;
for(int j=1;j<=min(i-1,len-i-1);++j){
if(s[i-j]==s[i+1+j]){
g[i-j][i+1+j]=1;
}else break;
}
}
}
Definition of F [i] [j] denotes the i to j answers this
有转移：$f[l][r]=f[l+1][r]+f[l][r-1]-f[l+1][r-1]+g[l][r]$
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define re register
using namespace std;
const int MAXN=5005;
char s[MAXN];
int t,len,f[MAXN][MAXN];
bool g[MAXN][MAXN];
signed main(){
scanf("%s",s+1);
len=strlen(s+1);
for(int i=1;i<=len;++i){
g[i][i]=1;
f[i][i]=1;
for(int j=1;j<=min(i-1,len-i);++j){
if(s[i-j]==s[i+j]){
g[i-j][i+j]=1;
}else break;
}
}
for(int i=1;i<=len-1;++i){
if(s[i]==s[i+1]){
g[i][i+1]=1;
f[i][i+1]=3;
for(int j=1;j<=min(i-1,len-i-1);++j){
if(s[i-j]==s[i+1+j]){
g[i-j][i+1+j]=1;
}else break;
}
}
}
for(int i=2;i<=len;++i){
for(int l=1;l<=len-i+1;++l){
int r=l+i-1;
f[l][r]=f[l+1][r]+f[l][r-1]-f[l+1][r-1]+g[l][r];
}
}
scanf("%lld",&t);
while(t--){
re int l,r;
scanf("%lld%lld",&l,&r);
printf("%lld\n",f[l][r]);
}
return 0;
}