Codeforces Round #266 (Div. 2)-c,d

C-number of Ways

Direct violence was sought in the past. Suppose you find the location of 1/3sum, then Mark + +. Find the 2/3 position, total plus the number of markers.

#include <stdio.h> #include <iostream> #include <stdlib.h> #include <string.h> #include < algorithm> #include <vector> #include <math.h> #include <queue> #include <stack> #include <map> #pragma COMMENT (linker, "/stack:1024000000,1024000000") using namespace std; #define MAXN 550000#define MoD    10000007#define LL __int64ll a[maxn];int main () {int n;        while (~SCANF ("%d", &n)) {LL sum=0;            for (int i=1;i<=n;i++) {scanf ("%i64d", &a[i]);        Sum+=a[i];            } if (sum%3) {cout<< "0" <<endl;        Continue        } SUM=SUM/3;        LL ans=0;        LL now=0;        LL s=0;            for (int i=1;i<n;i++) {now+=a[i];            if (Sum*2==now) {ans+=s;            } if (now==sum) {s++;    }} cout<<ans<<endl; } return 0;}

D-increase Sequence

First, change the An array into an H.

The b array is then derived from the forward difference to the a array.

CNT: Mark to current position, several L do not match R

Suppose B[i]==1:

Indicates the current position has a l,cnt++;

Suppose b[i]==0:

1, there's nothing in the current position.

2, the current position has an L, an R.

Because there is an L, so cnt++.

There is an R. So the total *=cnt,cnt--;

Equivalent to Total *= (cnt+1);

Suppose B[i]==-1:

The current position has an R, so the total *=cnt,cnt--;

Suppose B[i] is not equal to the above three cases, explain no solution!

#include <stdio.h> #include <iostream> #include <stdlib.h> #include <string.h> #include < algorithm> #include <vector> #include <math.h> #include <queue> #include <stack> #include <map> #pragma COMMENT (linker, "/stack:1024000000,1024000000") using namespace std; #define MAXN 550000#define MoD 1000000007#define LL __int64ll A[MAXN];    LL B[maxn];int Main () {int n,h;        while (~SCANF ("%d%d", &n,&h)) {for (int i=1;i<=n;i++) scanf ("%i64d", &a[i]);        for (int i=1;i<=n;i++) a[i]=h-a[i];        for (int i=1;i<=n+1;i++) {b[i]=a[i]-a[i-1];        } LL Ans=1;        LL cnt=0;            for (int i=1;i<=n+1;i++) {if (b[i]==1) {cnt++;                } else if (b[i]==-1) {ans=ans* (CNT)%mod;            cnt--;            } else if (b[i]==0) {ans=ans* (cnt+1)%mod;     } else       {ans=0;break;        } Ans=ans%mod;    } cout<<ans<<endl; } return 0;}

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Codeforces Round #266 (Div. 2)-c,d