Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 654 Accepted Submission(s): 219

#include<bits/stdc++.h>
using namespace std;
typedef __int64 INT;
const int maxn=55000;
int a[maxn],s[maxn];
INT cal(INT nn){
if(nn<3)
return 0;
return (nn-2)*(nn-1)*nn/6;
}
INT GCD(INT a,INT b){
return b==0?a:GCD(b,a%b);
}
int main(){
int t,n,m,li,ri;
scanf("%d",&t);
while(t--){
memset(s,0,sizeof(s));
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++){ //差分
scanf("%d%d",&li,&ri);
s[li]++;s[ri+1]--;
}
for(int i=1;i<=n;i++){ //前缀和。s数组中的值就是第i个玩具在多少个区间内。
s[i]+=s[i-1];
}
INT fm,fz;
fz=0;
for(int i=1;i<=n;i++){
fz+=cal((INT)s[i])*a[i];
}
if(m<3){
puts("0");
continue;
}
fm=cal(m);
if(fz==0){
printf("0\n",fm);
}else {
INT gcd=GCD(fz,fm);
fz/=gcd,fm/=gcd;
if(fm==1)
printf("%I64d\n",fz);
else
printf("%I64d/%I64d\n",fz,fm);
}
}
return 0;
}
HDU 5419——Victor and Toys——————【线段树|差分前缀和】
原文:http://www.cnblogs.com/chengsheng/p/4781365.html