dp题 并运用了前缀和
我看题目提示中有fft 我想了下感觉复杂度不过关还是未解
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 4e5+5;
const int MOD = 1e9+7;
const int ze = 2e5;
ll dp[2][MAXN];
ll sum[MAXN];
int main(){
int a,b,k,t;
while(~scanf("%d %d %d %d",&a,&b,&k,&t)) {
memset(dp,0,sizeof(dp));
int fl= 1;
dp[1][ze] = 1; sum[0] = 0;
for(int i = 1; i <= t; ++i) {
for(int j = 1; j < MAXN; ++j) {
sum[j] = (dp[fl][j]+sum[j-1]) %MOD;
}
memset(dp[fl^1],0,sizeof(dp[fl^1]));
for(int j = 1; j < MAXN; ++j) {
int t1 = max(0,j-k-1); int t2 = min(MAXN-1, j+k);
dp[fl^1][j] = (dp[fl^1][j]+sum[t2]-sum[t1]+MOD) %MOD;
}
fl ^= 1;
}
for(int i = 1; i < MAXN; ++i) {
sum[i] = (dp[fl][i] + sum[i-1]) %MOD;
}
ll ans = 0;
for(int i = 1; i < MAXN; ++i) {
int tt = a-b+i;
if(tt >= MAXN) tt = MAXN;
if(tt < 1) tt = 1;
ans = (ans + dp[fl][i]*sum[tt-1]%MOD)%MOD;
}
printf("%lld
",ans);
}
return 0;
}