1369 - Answering Queries<数>

E - Answering Queries
Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
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Description
The problem you need to solve here is pretty simple. You are give a function f(A, n), where A is an array of integers and n is the number of elements in the array. f(A, n) is defined as follows:


long long f( int A[], int n ) { // n = size of A
    long long sum = 0;
    for( int i = 0; i < n; i++ )
        for( int j = i + 1; j < n; j++ )
            sum += A[i] - A[j];
    return sum;
}
Given the array A and an integer n, and some queries of the form:


1)      0 x v (0 ≤ x < n, 0 ≤ v ≤ 106), meaning that you have to change the value of A[x] to v.
2)      1, meaning that you have to find f as described above.


Input
Input starts with an integer T (≤ 5), denoting the number of test cases.


Each case starts with a line containing two integers: n and q (1 ≤ n, q ≤ 105). The next line contains n space separated integers between 0 and 106 denoting the array A as described above.


Each of the next q lines contains one query as described above.


Output
For each case, print the case number in a single line first. Then for each query-type "1" print one single line containing the value of f(A, n).


Sample Input
1
3 5
1 2 3
1
0 0 3
1
0 2 1
1
Sample Output
Case 1:
-4
0

4

血一样的教训，，，，，以后light oj  只用long long  ,,,,打死也不用__int64

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL n,q,shu[101000];LL s,sum;int main(){int t;LL a,c,ca,b,i;scanf("%d",&t); //   int t;scanf("%d",&t);int a,b,c;for (ca=1;ca<=t;ca++){scanf("%lld%lld",&n,&q);memset(shu,0,sizeof(shu));s=0;sum=0;for (i=0;i<n;i++){scanf("%lld",&shu[i]);//s+=shu[i];sum=sum+(n-1-i)*shu[i]-i*shu[i];}printf("Case %lld:\n",ca);/*for (i=0;i<n-1;i++){s-=shu[i];sum=sum+(n-1-i)*shu[i]-s;}*/while (q--){scanf("%lld",&a);if (a==1)printf("%lld\n",sum);else{scanf("%lld%lld",&b,&c);a=c;c=c-shu[b];shu[b]=a;sum=sum+(n-1-b)*c-b*c;}}}return 0;}