http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4101

题意：数组a通过交换一对数字,得到了b数组,给出x=和y=和b数组,问有多少对l,r(l<=r)能满足条件

C++版本一

题解：规律+数学

1、；

2、；

3、，则；

4、，则不合法数据；

5、，，则不合法数据；

6、，，对于某一个位置i，可以推出应该交换的位置，其中，不一定在范围内，不一定在范围内；

7、对于每个再完成上述操作后，放进vector；

8、可以查询vector是否操作 ,当然直接询问;

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r;
ll x,y;
ll ans,cnt,flag,temp,X,Y;
ll b[N];
vector<int>v[M];
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%d",&t);
    while(t--){
        scanf("%d%lld%lld",&n,&x,&y);
        X=Y=0;
        for(int i=1;i<=100000;i++)
            v[i].clear();
        for(int i=1;i<=n;i++){
            scanf("%lld",&b[i]);
            X+=i*b[i];
            Y+=i*b[i]*b[i];
        }
        //cout<<X<<" "<<Y<<endl;
        ll needx=x-X,needy=y-Y;
        ans=0;
        if(needx==0&&needy==0){//needx==0&&needy==0时，任意两个相同的元素都可以交换
            for(int i=1;i<=n;i++){
                ans+=v[b[i]].size();
                v[b[i]].push_back(i);
            }
        }else if(needx==0){
            ans=0;
        }else if(needy%needx==0){//needx和needy必须倍数关系
            //cout<<needx<<" "<<needy<<endl;
            ll need=needy/needx;
            //cout<<need<<endl;
            for(int i=1;i<=n;i++){
                if(need-2*b[i]&&1<=need-b[i]&&need-b[i]<=100000){
                    int j=i-needx/(need-2*b[i]);
                    vector<int>::iterator it=lower_bound(v[need-b[i]].begin(),v[need-b[i]].end(),j);
                    //cout<<need-b[i]<<" "<<i-needx/(need-2*b[i])<<endl;
                    if(it!=v[need-b[i]].end()&&*it==j){//it不能是end(),不然会段错误
                        ans++;
                    }
                }
                v[b[i]].push_back(i);
            }
        }
        cout<<ans<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：

原博客 

#include<bits/stdc++.h>
#define ll long long 
#define Map map<ll,ll>::iterator
#define MAXN 100005
#define se second
using namespace std;
map<ll,ll>cnt;
ll X1,Y1,X2,Y2,a[MAXN];
int T,n;
int main(){
    cin>>T;
    while(T--){
        cnt.clear();
        scanf("%d%lld%lld",&n,&X1,&Y1);
        X2=Y2=0;
        for(ll i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            cnt[a[i]]++;
            X2+=a[i]*i;Y2+=a[i]*a[i]*i;
        }
        ll dx=X2-X1,dy=Y2-Y1;
        if(dx==0){
            if(dy!=0){
                puts("0");continue;
            }
            ll ans=0;
            for(Map it=cnt.begin();it!=cnt.end();it++){
                ans+=((it->se)-1)*(it->se)/2;
            }
            printf("%lld\n",ans);
            continue;
        }
        if(dy%dx){
            puts("0");continue;
        }
        //cout<<dx<<" "<<dy<<endl;
        ll dt=dy/dx,ans=0;
        //cout<<dt<<endl;
        for(ll i=1;i<=n;i++){
            ll Aj=dt-a[i];
            if((Aj-a[i])==0)continue;
            ll j=(dx+(Aj-a[i])*i)/(Aj-a[i]);
            if(j<=i||j>n)continue;
            if(a[j]==Aj)ans++;
        }
        printf("%lld\n",ans);
    }
}