cf----2019-09-22（Chocolates，Serval and Bus， Polycarp Restores Permutation）
明若清溪天下绝歌 缱绻成说，不知该在哪处着墨；一生情深怎奈何世事 徒留斑驳，只一念痴恋成奢。
You went to the store, selling nn types of chocolates. There are aiai chocolates of type ii in stock.
You have unlimited amount of cash (so you are not restricted by any prices) and want to buy as many chocolates as possible. However if you buy xixi chocolates of type ii (clearly, 0≤xi≤ai0≤xi≤ai), then for all 1≤j<i1≤j<i at least one of the following must hold:
- xj=0xj=0 (you bought zero chocolates of type jj)
- xj<xixj<xi (you bought less chocolates of type jj than of type ii)
For example, the array x=[0,0,1,2,10]x=[0,0,1,2,10] satisfies the requirement above (assuming that all ai≥xiai≥xi), while arrays x=[0,1,0]x=[0,1,0], x=[5,5]x=[5,5] and x=[3,2]x=[3,2] don't.
Calculate the maximum number of chocolates you can buy.
Input
The first line contains an integer nn (1≤n≤2⋅1051≤n≤2⋅105), denoting the number of types of chocolate.
The next line contains nn integers aiai (1≤ai≤1091≤ai≤109), denoting the number of chocolates of each type.
Output
Print the maximum number of chocolates you can buy.
Examples
input
Copy
5 1 2 1 3 6
output
Copy
10
input
Copy
5 3 2 5 4 10
output
Copy
20
input
Copy
4 1 1 1 1
output
Copy
1
Note
In the first example, it is optimal to buy: 0+0+1+3+60+0+1+3+6 chocolates.
In the second example, it is optimal to buy: 1+2+3+4+101+2+3+4+10 chocolates.
In the third example, it is optimal to buy: 0+0+0+10+0+0+1 chocolates.
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
ll a[200020];
int n;
int main()
{scanf ("%d",&n);for (int i=1; i<=n; i++)scanf ("%lld",&a[i]);ll ans=a[n];for (int i=n-1; i>=1; i--){if (a[i]>=a[i+1]){if (!a[i+1])break;ans+=a[i+1]-1;a[i]=a[i+1]-1;}elseans+=a[i];}printf ("%lld\n",ans);return 0;
}It is raining heavily. But this is the first day for Serval, who just became 3 years old, to go to the kindergarten. Unfortunately, he lives far from kindergarten, and his father is too busy to drive him there. The only choice for this poor little boy is to wait for a bus on this rainy day. Under such circumstances, the poor boy will use the first bus he sees no matter where it goes. If several buses come at the same time, he will choose one randomly.
Serval will go to the bus station at time tt, and there are nn bus routes which stop at this station. For the ii-th bus route, the first bus arrives at time sisi minutes, and each bus of this route comes didi minutes later than the previous one.
As Serval's best friend, you wonder which bus route will he get on. If several buses arrive at the same time, you can print any of them.
Input
The first line contains two space-separated integers nn and tt (1≤n≤1001≤n≤100, 1≤t≤1051≤t≤105) — the number of bus routes and the time Serval goes to the station.
Each of the next nn lines contains two space-separated integers sisi and didi (1≤si,di≤1051≤si,di≤105) — the time when the first bus of this route arrives and the interval between two buses of this route.
Output
Print one number — what bus route Serval will use. If there are several possible answers, you can print any of them.
Examples
input
Copy
2 2 6 4 9 5
output
Copy
1
input
Copy
5 5 3 3 2 5 5 6 4 9 6 1
output
Copy
3
input
Copy
3 7 2 2 2 3 2 4
output
Copy
1
Note
In the first example, the first bus of the first route arrives at time 66, and the first bus of the second route arrives at time 99, so the first route is the answer.
In the second example, a bus of the third route arrives at time 55, so it is the answer.
In the third example, buses of the first route come at times 22, 44, 66, 88, and so fourth, buses of the second route come at times 22, 55, 88, and so fourth and buses of the third route come at times 22, 66, 1010, and so on, so 11 and 22 are both acceptable answers while 33 is not.
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int main()
{int n, m,a,d, mi = inf, jg, ls;scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++){scanf("%d%d", &a, &d);if (a >= m)ls = a - m;else{ls = d - (m - a) % d;if (ls >= d)ls -= d;}if (ls < mi){mi = ls;jg = i;}}printf("%d\n", jg);return 0;
}An array of integers p1,p2,…,pnp1,p2,…,pn is called a permutation if it contains each number from 11 to nn exactly once. For example, the following arrays are permutations: [3,1,2][3,1,2], [1][1], [1,2,3,4,5][1,2,3,4,5] and [4,3,1,2][4,3,1,2]. The following arrays are not permutations: [2][2], [1,1][1,1], [2,3,4][2,3,4].
Polycarp invented a really cool permutation p1,p2,…,pnp1,p2,…,pn of length nn. It is very disappointing, but he forgot this permutation. He only remembers the array q1,q2,…,qn−1q1,q2,…,qn−1 of length n−1n−1, where qi=pi+1−piqi=pi+1−pi.
Given nn and q=q1,q2,…,qn−1q=q1,q2,…,qn−1, help Polycarp restore the invented permutation.
Input
The first line contains the integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the length of the permutation to restore. The second line contains n−1n−1 integers q1,q2,…,qn−1q1,q2,…,qn−1 (−n<qi<n−n<qi<n).
Output
Print the integer -1 if there is no such permutation of length nn which corresponds to the given array qq. Otherwise, if it exists, print p1,p2,…,pnp1,p2,…,pn. Print any such permutation if there are many of them.
Examples
input
Copy
3 -2 1
output
Copy
3 1 2
input
Copy
5 1 1 1 1
output
Copy
1 2 3 4 5
input
Copy
4 -1 2 2
output
Copy
-1
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int n,ans;
int a[200010];
int b[200010];
int p[200010];
int main()
{scanf("%d",&n);for(int i=1; i<n; i++){scanf("%d",&a[i]);p[i]=p[i-1]+a[i];}int mi=p[0];for(int i=1; i<n; i++)mi=min(mi,p[i]);for(int i=0; i<n; i++){if(p[i]-mi+1>n||b[p[i]-mi+1]){printf("-1\n");return 0;}b[p[i]-mi+1]=1;}for(int i=0; i<n; i++)printf("%d ",p[i]-mi+1);return 0;
}
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