技巧（1）sum[end] - sum[start]

 Flowers

Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %I64d      Java class name: Main

Submit Status PID: 26111

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times. 
In the next N lines, each line contains two integer Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, means the time of i-th query.

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

 

Sample Input

2
1 1
5 10
4
2 3
1 4
4 8
1
4
6

Sample Output

Case #1:
0
Case #2:
1
2
1

#include<cstdio>
#include<algorithm>
using namespace std;
int s[100005];
int e[100005];
int main(){
    int T,cas = 0;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++){
            scanf("%d%d",&s[i],&e[i]);
        }
        sort(s,s+n);
        sort(e,e+n);
        printf("Case #%d:\n",++cas);
        while(m--){
            int ans = 0;
            int x;
            scanf("%d",&x);
            ans += upper_bound(s,s+n,x)-s;//在时间x前开的花的数目，因为是从0开始标记的，所以他的下标不是他本身，而是他前面所有的数量；是upper_bound.
            ans -= lower_bound(e,e+n,x)-e;//在时间x前凋谢的花的数目，注意在时间t凋谢，但时间t时花还是开着的。所以是lower_bound.
            printf("%d\n",ans);
        }
    }
    return 0;
}

/*帮助理解的TLE代码*/
#include<cstdio>
#include<algorithm>
using namespace std;
int s[100005];
int e[100005];
int main(){
    int T,cas = 0;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++)
            scanf("%d%d",&s[i],&e[i]);
        sort(s,s+n);
        sort(e,e+n);
        while(m--){
            int cnt,x,ans = 0;
            scanf("%d",&x);
            for(cnt = 0; cnt < n; cnt++)
                if(s[cnt] > x)break;
            ans += cnt;
            for(cnt = 0; cnt < n; cnt++)
            if(e[cnt] >= x)break;
            ans -= cnt;
            printf("%d\n",ans);
        }
    }
    return 0;
}