Noip模拟题解题报告

Pro

题目链接

Sol

任务安排

读懂题之后果断想到二分，暴力判断就可以水过去。

听说贪心也可以……

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

struct Node {
	int time , end;
	bool operator < (const Node &a) const {
		return end < a.end;
	}
}; 
Node val[100005];
int n , l , r;

int jud(int x) {
	int res = x;
	for(int i=1; i<=n; i++) {
		if(res+val[i].time>val[i].end)
			return 0;
		res += val[i].time;
	}
	return 1;
}

int main() {
	freopen("manage.in","r",stdin);
	freopen("manage.out","w",stdout);
	scanf("%d",&n);
	for(int i=1; i<=n; i++)
		scanf("%d%d",&val[i].time,&val[i].end);
	sort(val+1 , val+n+1);
	l = 0;
	r = val[n].end;
	while(l<=r) {
		int mid = (l+r)>>1;
		if(jud(mid))
			l = mid + 1;
		else
			r = mid - 1;
	}
	printf("%d",r);
	return 0;
}

小 a 的强迫症

又是组合数……

公式见代码…… 不过预处理阶乘的时候0的位置放1会省去一些麻烦……

#include<iostream>
#include<cstdio>
#define mod 998244353
using namespace std;

long long n , num[100005] , sum = 0 , jc[500005] , ans = 1;

inline long long mypow(long long a , long long b) {
	if(!b)
		return 1;
	long long t = mypow(a , b>>1)%mod;
	if(b&1)
		return t*t%mod*a%mod;
	return t*t%mod;
}

inline long long inv(long long x) {
	return mypow(x , mod-2)%mod;
}

inline void init() {
	jc[1] = 1;
	for(int i=2; i<=sum; i++)
		jc[i] = jc[i-1]*i%mod;
}

inline long long C(long long x , long long y) {
	return jc[x]*inv(jc[y])%mod*inv(jc[x-y])%mod;
}

int main() {
	freopen("qiang.in","r",stdin);
	freopen("qiang.out","w",stdout);
	scanf("%lld",&n);
	for(int i=1; i<=n; i++) {
		scanf("%lld",&num[i]);
		sum += num[i];
	}
	init();
	for(int i=n; i>=1; i--) {
		long long t = C(sum-1,num[i]-1)%mod;
		ans = ans*(t==0?1:t)%mod;
		sum -= num[i];
	}
	printf("%lld",ans%mod);
	return 0;
} 

函数求和

最恶心的一道题…… 我打的暴力线段树 水了20分

正解是：分块+树状数组

暴力线段树

#include<iostream>
#include<cstdio>
using namespace std;

const int L = 100005;
struct Seg {
	long long l , r , sum;
};
Seg tree[4*L];
struct Node {
	long long l , r;
};
Node q[L];
long long n , data[L] , Q;

void build(long long num , long long l , long long r) {
	tree[num].l = l;
	tree[num].r = r;
	if(l == r) {
		tree[num].sum = data[l];
		return ;
	}
	long long mid = (l+r)>>1;
	build(num<<1 , l , mid);
	build(num<<1|1 , mid+1 , r);
	tree[num].sum = tree[num<<1].sum + tree[num<<1|1].sum;
}

void update(long long num , long long aim , long long cnt) {
	if(tree[num].l==tree[num].r&&tree[num].l==aim) {
		tree[num].sum = cnt;
		return ;
	}
	long long mid = (tree[num].l+tree[num].r)>>1;
	if(mid<aim)
		update(num<<1|1 , aim , cnt);
	else
		update(num<<1 , aim , cnt);
	tree[num].sum = tree[num<<1].sum + tree[num<<1|1].sum;
}

long long query(long long num , long long l , long long r) {
	if(l<=tree[num].l&&r>=tree[num].r)
		return tree[num].sum;
	long long mid = (tree[num].l+tree[num].r)>>1 , res = 0;
	if(l<=mid)
		res += query(num<<1 , l , r);
	if(r>mid)
		res += query(num<<1|1 , l , r);
	return res;
}

int main() {
	freopen("sum.in","r",stdin);
	freopen("sum.out","w",stdout);
	scanf("%lld",&n);
	for(int i=1; i<=n; i++)
		scanf("%lld",&data[i]);
	build(1 , 1 , n);
	for(int i=1; i<=n; i++)
		scanf("%lld%lld",&q[i].l,&q[i].r);
	scanf("%lld",&Q);
	while(Q--) {
		long long pot , x , y;
		scanf("%lld%lld%lld",&pot,&x,&y);
		if(pot == 1)
			update(1 , x , y);
		if(pot == 2) {
			long long ans = 0;
			for(int i=x; i<=y; i++)
				ans += query(1 , q[i].l , q[i].r);
			printf("%lld\n",ans);
		}
	}
	return 0;
}

分块+树状数组

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef unsigned long long LL;
#define lowbit(x) ((x)&-(x))
int n;
int a[100005];
LL c[100005];
int L[100005],R[100005];
int Q;
int t[405][100005];
int belong[100005];// (i-1)/B+1
LL sum[405];
int B,NUM;
void add(int x,LL d){
    while(x<=n){
        c[x]+=d;
        x+=lowbit(x);
    }
}
LL ask(int x){
    LL ret=0;
    while(x){
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}
int main(){
    freopen("sum.in","r",stdin);
    freopen("sum.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++){
        scanf("%d%d",&L[i],&R[i]);
    }
    B=sqrt(n)+1;
//  printf("#%d\n",B);
    NUM=(n-1)/B+1;
    for(int i=1;i<=n;i++) belong[i]=(i-1)/B+1;
    for(int i=1;i<=n;i++) c[i]=a[i];
    for(int i=1;i<=n;i++){
        if(i+lowbit(i)<=n){
            c[i+lowbit(i)]+=c[i];
        }
    }
    for(int i=1;i<=n;i++){
        t[belong[i]][L[i]]++;
        t[belong[i]][R[i]+1]--;
    }
    for(int i=1;i<=NUM;i++){
        for(int j=1;j<=n;j++){
            t[i][j]+=t[i][j-1];
            sum[i]+=t[i][j]*1ULL*a[j];
        }
    }
    scanf("%d",&Q);
    while(Q--){
        int type,x,y;
        scanf("%d%d%d",&type,&x,&y);
        if(type==1){
            for(int i=1;i<=NUM;i++){
                sum[i]-=t[i][x]*1ULL*a[x];
                sum[i]+=t[i][x]*1ULL*y;
            }
            add(x,-a[x]);
            a[x]=y;
            add(x,a[x]);
        }else{
            int Ln,Rn;
            Ln=belong[x],Rn=belong[y];
            LL ans=0;
            if(Ln==Rn){
                for(int i=x;i<=y;i++){
                    ans+=ask(R[i])-ask(L[i]-1);
                }
            }else{
                for(int i=Ln+1;i<Rn;i++) ans+=sum[i];
                int lim;
                lim=Ln*B;
                lim=min(lim,y);
                for(int i=x;i<=lim;i++){
                    ans+=ask(R[i])-ask(L[i]-1);
                }
                lim=(Rn-1)*B+1;
                lim=max(lim,x);
                for(int i=lim;i<=y;i++){
                    ans+=ask(R[i])-ask(L[i]-1);
                }
            }
            printf("%llu\n",ans);
        }
    }
    fclose(stdout);
    return 0;
}