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區間修改、單點查詢

Color the ball

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29305    Accepted Submission(s): 14264


Problem Description N個氣球排成一排,從左到右依次編號為1,2,3....N.每次給定2個整數a b(a <= b),lele便為騎上他的“小飛鴿"牌電動車從氣球a開始到氣球b依次給每個氣球塗一次顏色。但是N次以後lele已經忘記了第I個氣球已經塗過幾次顏色了,你能幫他算出每個氣球被塗過幾次顏色嗎?  

 

Input 每個測試例項第一行為一個整數N,(N <= 100000).接下來的N行,每行包括2個整數a b(1 <= a <= b <= N)。
當N = 0,輸入結束。  

 

Output 每個測試例項輸出一行,包括N個整數,第I個數代表第I個氣球總共被塗色的次數。  

 

Sample Input 3 1 1 2 2 3 3 3 1 1 1 2 1 3 0  

 

Sample Output 1 1 1 3 2 1  

 

Author 8600  

 

Source    
//hdu  1556
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
#include <map>
using namespace std;
#define ll long long
#define  lowbit(x) x&(-x)
const
int N=1e5+9; ll a[N],c[N]; ll n,x,y; void update(ll x,ll num){ while(x<=n){ c[x]+=num; x+=lowbit(x); } } ll getsum(ll x){ ll sum = 0; while(x>0){ sum+=c[x]; x-=lowbit(x); } return sum; } int main() { while(~scanf("%lld",&n),n){ for(ll i = 1 ;i <=n+9;i++){ a[i]=0; c[i] =0; } for(ll i =0;i<n;i++){ scanf("%lld%lld",&x,&y); update(x,1); update(y+1,-1); } for(ll i =1;i<=n;i++){ printf("%lld%c",getsum(i),i==n?'\n':' '); } } return 0; }