Codeforces Round #229 (Div. 2)(C,E待續)

In the first sample, Inna first chooses type "vertical segments", and then she makes segments with ends at :(0, 0),(0, 2); and, for example,(1, 0),(1, 1). If she paints horizontal segments, she will need at least 3 segments.

In the third sample it is important to note that Inna doesn't have the right to change the type of the segments during the game. That's why she will need 3 horizontal or 3 vertical segments to end the game.

用set儲存有多少個不同的列和行，輸出小的。

#include<iostream>
#include<cstdio>
#include<map>
#include<set>
using namespace std;
int main()
{
    int n,x,y;
    set<int> r,l;
    cin>>n;
    while(n--)
    {
        cin>>x>>y;
        r.insert(x);
        l.insert(y);
    }
    cout<<((r.size()>l.size())?l.size():r.size())<<endl;
    return 0;
}

3
1 1 2
2 2 3

4

1
2
5

-1

In the first sample, Dima and Inna play the first two notes at volume 1 (1 + 1 = 2, the condition holds), they should play the last note at volumes1 and2. Sereja's total joy equals:1·1 + 1·1 + 1·2 = 4.

題意：一首音樂有n段，每一段用兩種樂器演奏，要求兩種樂器音量都小於ai，和正好等於bi，使得它們的乘積最大，joy加上他們的乘積，入股不滿足要求，joy-1

思路：兩個數相差越小乘積越大，所以只要滿足要求下，joy+(bi/2)*(bi-bi/2);

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=100010;
typedef long long LL;
int D[maxn];
int main()
{
    LL n,ans=0,x;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>D[i];
    for(int i=0;i<n;i++)
    {
        cin>>x;
        if(x<=1||2*D[i]<x){ans--;continue;}
        ans+=(x/2)*(x-x/2);
    }
    cout<<ans<<endl;
    return 0;
}

10 3 3
1010100011
1 3
1 6
4 9

1
3
2

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
const int maxn=100010;
int n,k,w,l,r;
string a;
int ans[maxn],is_one[maxn];
void solve()
{
    memset(ans,0,sizeof(ans));
    memset(is_one,0,sizeof(is_one));
    int len=a.size();
    ans[0]=a[0]=='1'?1:0;
    for(int i=1;i<len;i++)
    ans[i]=ans[i-1]+(a[i]=='1'?1:0);
    for(int i=len-1;i>=len-k;i--)
    if(a[i]=='1')is_one[i]++;
    for(int i=len-k-1;i>=0;i--)
    {
        if(a[i]=='1')is_one[i]++;
        is_one[i]+=is_one[i+k];
    }
}
int main()
{
    cin>>n>>k>>w>>a;
    solve();
    while(w--)
    {
        cin>>l>>r;
        int q=(r+1-l)/k;
        l--,r--;
        int m=is_one[l+k-1]-is_one[r];
        if(a[r]=='1')m++;
        int p=l==0?ans[r]:ans[r]-ans[l-1];
        cout<<p-m+q-m<<endl;
    }
    return 0;
}

4 4 4

8
(1,1) (2,1) (2,2)
(1,1) (1,2)
(1,1) (2,1)
(1,1)

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=55;
struct node
{
    int x,y;
}grid[maxn*maxn];
int n,m,k,num;
bool cmp(node a,node b)
{
    return a.x+a.y<b.x+b.y;
}
int main()
{
    num=0;
    cin>>n>>m>>k;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        grid[num].x=i,grid[num++].y=j;
    sort(grid,grid+n*m,cmp);
    int ans=0;
    for(int i=0;i<k;i++)
    ans+=grid[i].x+grid[i].y-1;
    cout<<ans<<endl;
    for(int i=k-1;i>=0;i--)
    {
        for(int j=1;j<=grid[i].x;j++)
        printf("(%d,%d) ",j,1);
        for(int j=2;j<=grid[i].y;j++)
        printf("(%d,%d) ",grid[i].x,j);
        cout<<endl;
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 4 * 1000005;

int n[2], w;
ll bit1[2][N], bit2[2][N];

void insert(ll* b, int m, int x, ll value) {
	while (x <= m) {
		b[x] += value;
		x += x & (-x);
	}
}

ll get(ll* b, int x) {
	ll ans = 0;
	while (x > 0) {
		ans += b[x];
		x -= x & (-x);
	}
	return ans;
}

ll Get(int id, int r) {
	return get(bit1[id], r) * r + get(bit2[id], r);
}

void Insert(int id, int l, int r, ll value) {
	insert(bit1[id], n[id], l, value);
	insert(bit1[id], n[id], r+1, -value);
	insert(bit2[id], n[id], l, -value * (l-1));
	insert(bit2[id], n[id], r+1, value * r);
}

int main () {
	int x1, x2, y1, y2, v;
	scanf("%d%d%d", &n[0], &n[1], &w);
	while (w--) {
		scanf("%d%d%d%d%d", &v, &x1, &y1, &x2, &y2);
		if (v) {
			ll ans = Get(1, y2) - Get(1, y1-1) - (Get(0, n[0]) - Get(0, x2) + Get(0, x1-1));
			cout << ans << endl;
		} else {
			ll value;
			cin >> value;
			Insert(0, x1, x2, value * (y2-y1+1));
			Insert(1, y1, y2, value * (x2-x1+1));
		}
	}
	return 0;
}