题目链接：http://codeforces.com/problemset/problem/551/E
题意：给定一个长度为N的序列。 有2个操作 1 l r v:序列第l项到第r项加v(区间加)， 2 v：求整个序列中值为v的数的位置的差值最大是多少。不存在输出-1.
思路：分块。 每块维护该块序列排序后的序列。 对于区间修改，我们定义一个lazy标记。对于整块的修改我们只修改lazy, 其他情况暴力修改。然后情况该块后重新加入修改后的块然后排序。 对于查询操作。查询每块时v应该减去该块的lazy值。 然后2分查即可。
#define _CRT_SECURE_NO_DEPRECATE
#include<stdio.h>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<queue>
#include<math.h>
#include<time.h>
#include<vector>
#include<iostream>
#include<map>
using namespace std;
typedef long long int LL;
const int MAXN = * + ;
int belong[MAXN], block, num, L[MAXN], R[MAXN];
int n, q;
LL val[MAXN], lazy[MAXN];
struct Node{
LL v;
int id;
Node(LL _v, int _id) :v(_v), id(_id){};
bool operator < (const Node &a)const{
return v==a.v?id<a.id:v<a.v;
}
};
vector<Node>Bval[MAXN];
void build(){
block = (int)sqrt(n+0.5);
num = n / block; if (n%block){ num++; }
for (int i = ; i <= num; i++){
lazy[i] = ; Bval[i].clear();
L[i] = (i - )*block + ; R[i] = i*block;
}
R[num] = n;
for (int i = ; i <= n; i++){
belong[i] = ((i - ) / block) + ;
}
for (int i = ; i <= num; i++){
for (int k = L[i]; k <= R[i]; k++){
Bval[i].push_back(Node(val[k],k));
}
sort(Bval[i].begin(), Bval[i].end());
}
}
void modify(int st,int ed, LL v){
if (belong[st] == belong[ed]){
for (int i = st; i <= ed; i++){
val[i] += v;
}
Bval[belong[st]].clear();
for (int i = L[belong[st]]; i <= R[belong[st]]; i++){
Bval[belong[st]].push_back(Node(val[i], i));
}
sort(Bval[belong[st]].begin(), Bval[belong[st]].end());
return;
}
for (int i = st; i <= R[belong[st]]; i++){
val[i] += v;
}
Bval[belong[st]].clear();
for (int i = L[belong[st]]; i <= R[belong[st]]; i++){
Bval[belong[st]].push_back(Node(val[i], i));
}
sort(Bval[belong[st]].begin(), Bval[belong[st]].end());
for (int i = belong[st] + ; i < belong[ed]; i++){
lazy[i] += v;
}
for (int i = L[belong[ed]]; i <= ed; i++){
val[i] += v;
}
Bval[belong[ed]].clear();
for (int i = L[belong[ed]]; i <= R[belong[ed]]; i++){
Bval[belong[ed]].push_back(Node(val[i], i));
}
sort(Bval[belong[ed]].begin(), Bval[belong[ed]].end());
}
int query(LL v){
int L = -, R = -;
for (int i = ; i <= num; i++){
LL _v = v-lazy[i];
int pos = lower_bound(Bval[i].begin(), Bval[i].end(), Node(_v,)) - Bval[i].begin();
if (pos >= && pos < Bval[i].size() && Bval[i][pos].v == _v){
L = Bval[i][pos].id; break;
}
}
if (L == -){ return -; }
for (int i = num; i > ; i--){
LL _v = v - lazy[i];
int pos = (lower_bound(Bval[i].begin(), Bval[i].end(), Node( _v + ,)) - Bval[i].begin())-;
if (pos >= && pos < Bval[i].size() && Bval[i][pos].v == _v){
R = Bval[i][pos].id; break;
}
}
return R - L;
}
int main(){
//#ifdef kirito
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
//#endif
// int start = clock();
while (~scanf("%d%d", &n,&q)){
for (int i = ; i <= n; i++){
scanf("%lld", &val[i]);
}
build();
for (int i = ; i <= q; i++){
int type, l, r, v;
scanf("%d", &type);
if (type == ){
scanf("%d%d%lld", &l, &r, &v);
modify(l, r, v);
}
else{
scanf("%lld", &v);
printf("%d\n", query(v));
}
}
}
//#ifdef LOCAL_TIME
// cout << "[Finished in " << clock() - start << " ms]" << endl;
//#endif
return ;
}