二分类型。

1.输出数组中大于等于 l 且小于等于 r 的元素个数 

法一： 

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,m;
    cin>>n>>m;
   vector<int>q(n);
    for(int i=0;i<n;i++){
        scanf("%d",&q[i]);
    }
    sort(q.begin(),q.end());
  while(m--){
        int l,r;
        cin>>l>>r;
        int left=0,right=n-1;
      
        while(left<right){
            int mid=right+left>>1;
            if(q[mid]>=l)   right=mid;             
            else left=mid+1;              
        }
        int res1=left;
      
        left=0,right=n-1;
        while(left<right){
            int mid=(left+right+1)/2;//防止陷入死循环
            if(q[mid]<=r) left=mid;
            else right=mid-1;
        }
        int res2=right;
      
        cout<<res2-res1+1<<endl;
    }
    return 0;
}

法二： 

#include<bits/stdc++.h>
using namespace std; 
int n,q,l,r;
int main()
{
    cin>>n>>q;
    vector<int>arr(n);
    for(int i=0;i<n;i++)
        cin>>arr[i];
    std::sort(arr.begin(),arr.end());
    while(q--)
    {
        cin>>l>>r;
        auto left=lower_bound(arr.begin(),arr.end(),l);
        auto right=upper_bound(arr.begin(),arr.end(),r);
        cout<<right-left<<endl;
    }
    return 0;
}

lower_bound:查找第一次大于等于l的 元素位置

upper_bound:查找第一次大于r的元素位置

2.音符

法一：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,q;
    cin >> n >> q;
    vector<int> a(n+1,0);
    vector<int> c(n+1,0);
    for(int i=1; i<=n; i++)
    {
        cin >> a[i];
        c[i] = c[i-1] + a[i];
    }
    while(q--)
    {
        int b;
        cin >> b;
        int l = 1, r = n;
        while(l < r)
        {
            int mid = (l + r ) >> 1;
            if(c[mid] > b) r = mid;
            else l = mid+1;
        }
        int tmp = l;
        cout << tmp << endl;
    }
    return 0;
}

法二：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,q;
    cin >> n >> q;
    vector<int> a(n+1,0);
    vector<int> c(n+1,0);
    for(int i=1; i<=n; i++)
    {
        cin >> a[i];
        c[i] = c[i-1] + a[i];
    }
    while(q--)
    {
        int b;
        cin >> b;
        auto it = upper_bound(c.begin()+1, c.end(), b);
        int tmp = it - c.begin();
        cout << tmp << endl;
    }
    return 0;
}

3.洛谷p1918保龄球

法一：二分递归

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,a,ans;
struct E{
	ll h,H;
}N[100005];
bool cmp(E x,E y)
{
	return x.h<y.h;
}
bool pd(ll A,ll l,ll r)
{
	ll mid=(l+r)/2;
	if(N[mid].h==A)
	{
		ans=N[mid].H;
		return true; 
	}
	if(l>r)return false;
	if(A<=N[mid].h)return pd(A,l,mid-1);
	else return pd(A,mid+1,r);
}
int main()
{
	cin>>n;
	for(ll i=1;i<=n;i++)
	{
		cin>>N[i].h;
		N[i].H=i;
	}
	sort(N+1,N+1+n,cmp);
	cin>>m;
	for(ll i=1;i<=m;i++)
	{
		cin>>a;
		if(pd(a,1,n))
		{
			cout<<ans<<endl;
		}
		else cout<<0<<endl;
	}
	return 0;
} 

法二：二分循环

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,a,ans;
struct E{
	ll h,H;
}N[100005];
bool cmp(E x,E y)
{
	return x.h<y.h;
}
ll pd(ll tar,ll l,ll r)
{
	while(l <= r){
	ll mid=(l+r)/2;
	if(N[mid].h==tar)
	{
		ans=N[mid].H;
		return ans; 
	}
	if(N[mid].h < tar) l = mid+1;
	if(N[mid].h > tar) r = mid-1;
   }
   return -1;
}
int main()
{
	cin>>n;
	for(ll i=1;i<=n;i++)
	{
		cin>>N[i].h;
		N[i].H=i;
	}
	sort(N+1,N+1+n,cmp);
	cin>>m;
	for(ll i=1;i<=m;i++)
	{
		cin>>a;
		ll tmp = pd(a,1,n);
		if(tmp == -1) cout << 0 << endl;
		else cout<< tmp <<endl;
	}
	return 0;
} 

法三：map+暴力

#include<map>
#include<iostream>
using namespace std;
map<int,int>ma;
int n,a,q,m;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a;
        ma[a]=i;
    }
    cin>>q;
    for(int i=1;i<=q;i++){
        cin>>m;
        cout<<ma[m]<<endl;
    }
}

4.洛谷P1163

#include <bits/stdc++.h>
#define up(l,r,i) for(int i=l;i<=r;i++)
#define dn(l,r,i) for(int i=r; i>=l;i--)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;
double n, m, s, ans;
int k;
int right(double u){
	double sum = 0, v = 1;
	up(1,k,i){
		v *= (1+u);
		sum += (m/v);
	}
	if(sum == n) return 0;
	if(sum > n) return 1;
	else return -1;
}
int main()
{
	IOS;
	cin >> n >> m >> k;
	double l=0, r=800, mid;
	while(r-l >= 0.0001){
		mid = (l+r) / 2;
		if(!right(mid)) break;
		if(right(mid) == 1) l = mid;
		else r = mid; 
	}
	printf("%.1lf", mid*100);
    return 0;
}

5.洛谷P2678

#include <bits/stdc++.h>
#define up(l,r,i) for(int i=l;i<=r;i++)
#define dn(l,r,i) for(int i=r; i>=l;i--)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int MAXN = 50001;
int stone[MAXN], a, n, m;
bool check(int d){
	int p=0, ans=0;
	up(1,n,i){
		if(stone[i] - p < d) ans++;
		else p = stone[i];
	}
	if(ans <= m) return true;
	else return false;
}
int main()
{
	IOS;
	cin >> a >> n >> m;
	up(1,n,i) cin >> stone[i];
	stone[++n] = a;
	int l = 0, r = a, mid;
	while(l < r){
		mid = (l+r+1) / 2;
		if(check(mid)) l = mid;
		else r = mid - 1;
	}
	cout << l ;
    return 0;
}

6.洛谷p1024

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define int long long
#define up(l,r,i) for(int i=l; i<=r; ++i)
using namespace std;
double a, b, c, d;
double fc(double x)
{
	return a*x*x*x+b*x*x+c*x+d;
}
signed main()
{
	IOS;
	double l,r,m,x1,x2;
	int s = 0, i;
	cin >> a >> b >> c >> d;
	up(-100,99,i){
		l = i;
		r = i+1;
		x1 = fc(l);
		x2 = fc(r);
		if(!x1)
		{
			printf("%.2lf ",l);
			s++;
		}
		if(x1 * x2 < 0)
		{
			while(r-l >= 0.001)
			{
				m = (l+r) / 2;
				if(fc(m)*fc(r) <= 0)l=m;
				else r=m;
			}
			printf("%.2lf ",r);
			s++;
		}
		if(s == 3) break;
	}
	return 0;
}