codeforces 1217E E. Sum Queries? (线段树

时间:2023-03-09 00:24:32
codeforces 1217E  E. Sum Queries? (线段树

codeforces 1217E E. Sum Queries? (线段树

传送门:https://codeforces.com/contest/1217/problem/E

题意:

n个数,m次询问

单点修改

询问区间内最小的unbalanced number

balanced number定义是,区间内选取数字的和sum

sum上的每一位都对应着选取的数上的一位

否则就是unbalanced number

题解:

根据题意

如果区间存在unbalance number,那么一定存在两个数就可以组成unbalance number

为什么?

因为根据balanced number的定义来说,只要在和的一位上 有两个数字在这个位置上的数都不为0的话那么这两个数字一定可以组成unbalanced number,所以我们需要对每一位非0的数进行操作

值域最多为2e9,所以我们建立十颗线段树,保存每一位上的 如果当前位不为0,保存这个位置上所在数的最小值

记录一个保存答案的线段树,答案就为 当前位的两个数都不为0的两数之和的最小值

代码:

/**
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*        ┃       ┃  
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*        ┃       ┃
*        ┃... ⌒ ...  ┃
*        ┃       ┃
*        ┗━┓   ┏━┛
*          ┃   ┃ Code is far away from bug with the animal protecting          
*          ┃   ┃ 神兽保佑,代码无bug
*          ┃   ┃           
*          ┃   ┃       
*          ┃   ┃
*          ┃   ┃           
*          ┃   ┗━━━┓
*          ┃       ┣┓
*          ┃       ┏┛
*          ┗┓┓┏━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 2e9+7;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
LL a[maxn];
LL Min[maxn << 2][15];
LL ANS[maxn << 2];
void push_up(int rt) {
ANS[rt] = INF;
for(int i = 0; i <= 12; i++) {
if(Min[ls][i] != INF && Min[rs][i] != INF) {
ANS[rt] = min(ANS[rt], Min[ls][i] + Min[rs][i]);
}
Min[rt][i] = min(Min[ls][i], Min[rs][i]);
}
ANS[rt] = min(ANS[rt], min(ANS[ls], ANS[rs]));
}
void build(int l, int r, int rt) {
ANS[rt] = INF;
if(l == r) {
int tmp = a[l];
// debug1(tmp);
for(int i = 0; i <= 12; i++) {
int val = tmp % 10;
if(val == 0) Min[rt][i] = INF;
else Min[rt][i] = a[l];
tmp /= 10;
}
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
push_up(rt);
}
void update(int pos, int x, int l, int r, int rt) {
if(l == r) {
int tmp = x;
ANS[rt] = INF;
// debug1(tmp);
for(int i = 0; i <= 12; i++) {
int val = tmp % 10;
if(val == 0) Min[rt][i] = INF;
else Min[rt][i] = x;
// Min[rt][i] = val;
// debug2(val,Min[rt][i]);
tmp /= 10;
}
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(pos, x, lson);
else update(pos, x, rson);
push_up(rt);
}
LL res[15];
LL ans = INF;
void query(int L, int R, LL &ans, int l, int r, int rt) {
// if(l > R || r < L) return;
if(L <= l && r <= R) {
for(int i = 0; i <= 12; i++) {
if(res[i] != INF && Min[rt][i] != INF) {
ans = min(ans, res[i] + Min[rt][i]);
}
}
for(int i = 0; i <= 10; i++) {
res[i] = min(res[i], Min[rt][i]);
}
ans = min(ANS[rt], ans);
return;
}
int mid = (l + r) >> 1;
if(L <= mid) query(L, R, ans, lson);
if(R > mid) query(L, R, ans, rson);
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build(1, n, 1);
while(m--) {
int op, l, r, pos, x;
scanf("%d", &op);
if(op == 1) {
scanf("%d%d", &pos, &x);
update(pos, x, 1, n, 1);
} else {
scanf("%d%d", &l, &r);
ans = INF;
for(int i = 0; i <= 12; i++) res[i] = INF;
query(l, r, ans, 1, n, 1);
if(ans == INF) ans = -1;
printf("%lld\n", ans);
}
}
return 0;
}