NYOJ 228 士兵杀敌（五）

http://acm.nyist.net/JudgeOnline/problem.php?pid=228

本想用树状数组的, 结果有点囧...........

然后看了别的大牛的做法....  离线查询的可以有o(n)的算法...

 
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn = 1000005;
const int M = 10003;
int ans[maxn];
int main()
{
    int n, m, q, a, b, c, i;
    scanf("%d%d%d",&n,&m,&q);
    while(m--)
    {
        scanf("%d%d%d",&a,&b,&c);
        ans[a]+=c;
        ans[b+1]-=c;
    }
    for(i=1; i<=n; i++)
    ans[i] += ans[i-1];
    for(i=1; i<=n; i++)
    ans[i]=(ans[i-1]+ans[i])%M;
    while(q--)
    {
        scanf("%d%d",&a,&b);
        printf("%d\n",(ans[b]-ans[a-1]+M)%M);
    }
    return 0;
}